Why does the Binomial Theorem use combinations and not permutations for its coefficients?












2












$begingroup$


I have been trying to understand the Binomial Theorem formula. I can see that it works.



What I don’t understand is how or why using combinations finds the coefficients.




What I mean is, isn’t each coefficient actually a permutation?




In the sense, that a combination isn’t concerned with the order. Yet the coefficient seems to reflect the ways a selection of items can be ordered.



It seems like a contradiction.



A simple explanation would be greatly appreciated. As I am not a mathematician.



Many thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Short answer: commutativity (and I guess associativity) of multiplication
    $endgroup$
    – Brevan Ellefsen
    Feb 21 at 9:41










  • $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
    $endgroup$
    – Jor
    Feb 22 at 21:44












  • $begingroup$
    my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:06












  • $begingroup$
    Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
    $endgroup$
    – Jor
    Feb 22 at 22:15










  • $begingroup$
    I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:20


















2












$begingroup$


I have been trying to understand the Binomial Theorem formula. I can see that it works.



What I don’t understand is how or why using combinations finds the coefficients.




What I mean is, isn’t each coefficient actually a permutation?




In the sense, that a combination isn’t concerned with the order. Yet the coefficient seems to reflect the ways a selection of items can be ordered.



It seems like a contradiction.



A simple explanation would be greatly appreciated. As I am not a mathematician.



Many thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Short answer: commutativity (and I guess associativity) of multiplication
    $endgroup$
    – Brevan Ellefsen
    Feb 21 at 9:41










  • $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
    $endgroup$
    – Jor
    Feb 22 at 21:44












  • $begingroup$
    my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:06












  • $begingroup$
    Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
    $endgroup$
    – Jor
    Feb 22 at 22:15










  • $begingroup$
    I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:20
















2












2








2


2



$begingroup$


I have been trying to understand the Binomial Theorem formula. I can see that it works.



What I don’t understand is how or why using combinations finds the coefficients.




What I mean is, isn’t each coefficient actually a permutation?




In the sense, that a combination isn’t concerned with the order. Yet the coefficient seems to reflect the ways a selection of items can be ordered.



It seems like a contradiction.



A simple explanation would be greatly appreciated. As I am not a mathematician.



Many thanks.










share|cite|improve this question











$endgroup$




I have been trying to understand the Binomial Theorem formula. I can see that it works.



What I don’t understand is how or why using combinations finds the coefficients.




What I mean is, isn’t each coefficient actually a permutation?




In the sense, that a combination isn’t concerned with the order. Yet the coefficient seems to reflect the ways a selection of items can be ordered.



It seems like a contradiction.



A simple explanation would be greatly appreciated. As I am not a mathematician.



Many thanks.







binomial-coefficients binomial-theorem binomial-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 21 at 8:53









Blue

48.6k870156




48.6k870156










asked Feb 21 at 8:30









JorJor

111




111












  • $begingroup$
    Short answer: commutativity (and I guess associativity) of multiplication
    $endgroup$
    – Brevan Ellefsen
    Feb 21 at 9:41










  • $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
    $endgroup$
    – Jor
    Feb 22 at 21:44












  • $begingroup$
    my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:06












  • $begingroup$
    Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
    $endgroup$
    – Jor
    Feb 22 at 22:15










  • $begingroup$
    I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:20




















  • $begingroup$
    Short answer: commutativity (and I guess associativity) of multiplication
    $endgroup$
    – Brevan Ellefsen
    Feb 21 at 9:41










  • $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
    $endgroup$
    – Jor
    Feb 22 at 21:44












  • $begingroup$
    my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:06












  • $begingroup$
    Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
    $endgroup$
    – Jor
    Feb 22 at 22:15










  • $begingroup$
    I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
    $endgroup$
    – Brevan Ellefsen
    Feb 22 at 22:20


















$begingroup$
Short answer: commutativity (and I guess associativity) of multiplication
$endgroup$
– Brevan Ellefsen
Feb 21 at 9:41




$begingroup$
Short answer: commutativity (and I guess associativity) of multiplication
$endgroup$
– Brevan Ellefsen
Feb 21 at 9:41












$begingroup$
I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
$endgroup$
– Jor
Feb 22 at 21:44






$begingroup$
I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing?
$endgroup$
– Jor
Feb 22 at 21:44














$begingroup$
my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
$endgroup$
– Brevan Ellefsen
Feb 22 at 22:06






$begingroup$
my comment is precisely what explains your comment. If none of $abb, bab,$ and $bba$ were equal (so multiplication is not at all commutative) then you would have permutations. Since they are all equal (by the commutativity of multiplication) we get combinations. If multiplication were only somewhat commutative (e.g. $bab = abb$ but $bab neq abb$) then we would get something between combinations and permutations.
$endgroup$
– Brevan Ellefsen
Feb 22 at 22:06














$begingroup$
Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
$endgroup$
– Jor
Feb 22 at 22:15




$begingroup$
Ok. Thanks. However, I don’t understand what I am missing please? In the binomial theorem, using the combination formula, the selection, (abb, bab and bba), turns out to be 3. Yet, clearly, it’s one combination..
$endgroup$
– Jor
Feb 22 at 22:15












$begingroup$
I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
$endgroup$
– Brevan Ellefsen
Feb 22 at 22:20






$begingroup$
I have no idea what you're talking about. The binomial theorem doesn't say anything like that; it's a way of expressing $(a+b) ^n$ as a sum of simpler terms. Could you give more clarification please?
$endgroup$
– Brevan Ellefsen
Feb 22 at 22:20












2 Answers
2






active

oldest

votes


















7












$begingroup$

The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.



Consider



$$(a+b)^3$$



If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see



$$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$



Notice that we can characterize the sum this way:



$$(a+b)^3 = (text{terms with 3 a's}) + (text{terms with 2 a's}) + (text{terms with 1 a}) + (text{terms with no a's})$$



(You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.



So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.



So we can conclude that



$$begin{align}
(text{# of terms with 3 a's}) &= binom{3}{3} = 1\
(text{# of terms with 2 a's}) &= binom{3}{2} = 3\
(text{# of terms with 1 a}) &= binom{3}{1} = 3\
(text{# of terms with no a's}) &= binom{3}{0} = 1
end{align}$$



Thus, we conclude:




  • There will only be one $aaa = a^3$ term

  • There will be $3$ $aba=aab=baa=a^2b$ terms.

  • There will be $3$ $abb = bab = abb = ab^2$ terms.

  • There will be $1$ $bbb=b^3$ term.


Thus,



$$(a+b)^3 = sum_{k=0}^3 binom{3}{k}a^k b^{3-k}$$



and in general, for positive integers $n$,



$$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$





In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).



Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).



The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
    $endgroup$
    – Jor
    Feb 22 at 21:48












  • $begingroup$
    For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
    $endgroup$
    – Jor
    Feb 22 at 21:57












  • $begingroup$
    It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
    $endgroup$
    – Jor
    Feb 22 at 22:03










  • $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56










  • $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56



















0












$begingroup$

My way of seeing the binomial formula is the following. Suppose you want to compute
$$
(a+b)^n
$$

for some $ngeq 1$. Look at it in this way:
$$
underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}},
$$

with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.






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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.



    Consider



    $$(a+b)^3$$



    If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see



    $$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$



    Notice that we can characterize the sum this way:



    $$(a+b)^3 = (text{terms with 3 a's}) + (text{terms with 2 a's}) + (text{terms with 1 a}) + (text{terms with no a's})$$



    (You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.



    So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.



    So we can conclude that



    $$begin{align}
    (text{# of terms with 3 a's}) &= binom{3}{3} = 1\
    (text{# of terms with 2 a's}) &= binom{3}{2} = 3\
    (text{# of terms with 1 a}) &= binom{3}{1} = 3\
    (text{# of terms with no a's}) &= binom{3}{0} = 1
    end{align}$$



    Thus, we conclude:




    • There will only be one $aaa = a^3$ term

    • There will be $3$ $aba=aab=baa=a^2b$ terms.

    • There will be $3$ $abb = bab = abb = ab^2$ terms.

    • There will be $1$ $bbb=b^3$ term.


    Thus,



    $$(a+b)^3 = sum_{k=0}^3 binom{3}{k}a^k b^{3-k}$$



    and in general, for positive integers $n$,



    $$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$





    In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).



    Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).



    The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
      $endgroup$
      – Jor
      Feb 22 at 21:48












    • $begingroup$
      For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
      $endgroup$
      – Jor
      Feb 22 at 21:57












    • $begingroup$
      It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
      $endgroup$
      – Jor
      Feb 22 at 22:03










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56
















    7












    $begingroup$

    The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.



    Consider



    $$(a+b)^3$$



    If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see



    $$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$



    Notice that we can characterize the sum this way:



    $$(a+b)^3 = (text{terms with 3 a's}) + (text{terms with 2 a's}) + (text{terms with 1 a}) + (text{terms with no a's})$$



    (You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.



    So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.



    So we can conclude that



    $$begin{align}
    (text{# of terms with 3 a's}) &= binom{3}{3} = 1\
    (text{# of terms with 2 a's}) &= binom{3}{2} = 3\
    (text{# of terms with 1 a}) &= binom{3}{1} = 3\
    (text{# of terms with no a's}) &= binom{3}{0} = 1
    end{align}$$



    Thus, we conclude:




    • There will only be one $aaa = a^3$ term

    • There will be $3$ $aba=aab=baa=a^2b$ terms.

    • There will be $3$ $abb = bab = abb = ab^2$ terms.

    • There will be $1$ $bbb=b^3$ term.


    Thus,



    $$(a+b)^3 = sum_{k=0}^3 binom{3}{k}a^k b^{3-k}$$



    and in general, for positive integers $n$,



    $$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$





    In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).



    Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).



    The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
      $endgroup$
      – Jor
      Feb 22 at 21:48












    • $begingroup$
      For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
      $endgroup$
      – Jor
      Feb 22 at 21:57












    • $begingroup$
      It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
      $endgroup$
      – Jor
      Feb 22 at 22:03










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56














    7












    7








    7





    $begingroup$

    The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.



    Consider



    $$(a+b)^3$$



    If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see



    $$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$



    Notice that we can characterize the sum this way:



    $$(a+b)^3 = (text{terms with 3 a's}) + (text{terms with 2 a's}) + (text{terms with 1 a}) + (text{terms with no a's})$$



    (You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.



    So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.



    So we can conclude that



    $$begin{align}
    (text{# of terms with 3 a's}) &= binom{3}{3} = 1\
    (text{# of terms with 2 a's}) &= binom{3}{2} = 3\
    (text{# of terms with 1 a}) &= binom{3}{1} = 3\
    (text{# of terms with no a's}) &= binom{3}{0} = 1
    end{align}$$



    Thus, we conclude:




    • There will only be one $aaa = a^3$ term

    • There will be $3$ $aba=aab=baa=a^2b$ terms.

    • There will be $3$ $abb = bab = abb = ab^2$ terms.

    • There will be $1$ $bbb=b^3$ term.


    Thus,



    $$(a+b)^3 = sum_{k=0}^3 binom{3}{k}a^k b^{3-k}$$



    and in general, for positive integers $n$,



    $$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$





    In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).



    Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).



    The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.






    share|cite|improve this answer











    $endgroup$



    The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.



    Consider



    $$(a+b)^3$$



    If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see



    $$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$



    Notice that we can characterize the sum this way:



    $$(a+b)^3 = (text{terms with 3 a's}) + (text{terms with 2 a's}) + (text{terms with 1 a}) + (text{terms with no a's})$$



    (You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.



    So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.



    So we can conclude that



    $$begin{align}
    (text{# of terms with 3 a's}) &= binom{3}{3} = 1\
    (text{# of terms with 2 a's}) &= binom{3}{2} = 3\
    (text{# of terms with 1 a}) &= binom{3}{1} = 3\
    (text{# of terms with no a's}) &= binom{3}{0} = 1
    end{align}$$



    Thus, we conclude:




    • There will only be one $aaa = a^3$ term

    • There will be $3$ $aba=aab=baa=a^2b$ terms.

    • There will be $3$ $abb = bab = abb = ab^2$ terms.

    • There will be $1$ $bbb=b^3$ term.


    Thus,



    $$(a+b)^3 = sum_{k=0}^3 binom{3}{k}a^k b^{3-k}$$



    and in general, for positive integers $n$,



    $$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$





    In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).



    Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).



    The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 21 at 8:52

























    answered Feb 21 at 8:46









    Eevee TrainerEevee Trainer

    7,00811337




    7,00811337












    • $begingroup$
      I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
      $endgroup$
      – Jor
      Feb 22 at 21:48












    • $begingroup$
      For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
      $endgroup$
      – Jor
      Feb 22 at 21:57












    • $begingroup$
      It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
      $endgroup$
      – Jor
      Feb 22 at 22:03










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56


















    • $begingroup$
      I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
      $endgroup$
      – Jor
      Feb 22 at 21:48












    • $begingroup$
      For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
      $endgroup$
      – Jor
      Feb 22 at 21:57












    • $begingroup$
      It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
      $endgroup$
      – Jor
      Feb 22 at 22:03










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56










    • $begingroup$
      "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
      $endgroup$
      – Eevee Trainer
      Feb 22 at 23:56
















    $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
    $endgroup$
    – Jor
    Feb 22 at 21:48






    $begingroup$
    I appreciate your answer. Thanks. However, I still don’t understand why combinations work. For example ab squared = abb, bab, or bba. So thats three versions of the same thing. So it’s a permutation... what am I missing please? If I imagine each set of brackets of (a+b) in different colours, if that makes sense, so that each selection is actually a new combination, then I can see it, but I’m not sure.
    $endgroup$
    – Jor
    Feb 22 at 21:48














    $begingroup$
    For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
    $endgroup$
    – Jor
    Feb 22 at 21:57






    $begingroup$
    For example, a yellow box with a yellow (a+b) in it, and a blue box with a blue (a+b) in it, and red box with a red (a+b) in it. Then, it would make sense to me, as there would be many more ways to arrange the coloured a’s and b’s, so way more permutations, but naturally less combinations. But again I’m uncertain if thinking this way is helpful. I find it easier to visualise the problem..
    $endgroup$
    – Jor
    Feb 22 at 21:57














    $begingroup$
    It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
    $endgroup$
    – Jor
    Feb 22 at 22:03




    $begingroup$
    It’s almost like combinations, used in this manner, is like saying... how many ways can I choose something. But why? When the definition of a combination, isn’t concerned with the numbers of ways of choosing... is it?
    $endgroup$
    – Jor
    Feb 22 at 22:03












    $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56




    $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56












    $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56




    $begingroup$
    "So thats three versions of the same thing. So it’s a permutation... what am I missing please?" - It's not a permutation, because the order doesn't matter. $aab, aba, baa$: all three of those are equivalent expressions, and there are $binom 3 2$ ways to write a sequence of length three, with only $a$ and $b$, that has precisely two $a$'s in it. The expansion of $(a+b)^3$ requires us to take an $a$ or $b$ from each $(a+b)$ term in the expression $(a+b)(a+b)(a+b)$ and that's also how we know we get every possible sequence of length three.
    $endgroup$
    – Eevee Trainer
    Feb 22 at 23:56











    0












    $begingroup$

    My way of seeing the binomial formula is the following. Suppose you want to compute
    $$
    (a+b)^n
    $$

    for some $ngeq 1$. Look at it in this way:
    $$
    underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}},
    $$

    with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      My way of seeing the binomial formula is the following. Suppose you want to compute
      $$
      (a+b)^n
      $$

      for some $ngeq 1$. Look at it in this way:
      $$
      underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}},
      $$

      with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        My way of seeing the binomial formula is the following. Suppose you want to compute
        $$
        (a+b)^n
        $$

        for some $ngeq 1$. Look at it in this way:
        $$
        underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}},
        $$

        with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.






        share|cite|improve this answer











        $endgroup$



        My way of seeing the binomial formula is the following. Suppose you want to compute
        $$
        (a+b)^n
        $$

        for some $ngeq 1$. Look at it in this way:
        $$
        underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}},
        $$

        with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 21 at 9:44

























        answered Feb 21 at 9:29









        AlessioDVAlessioDV

        655114




        655114






























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