How many combinations of coins add up to $20












2












$begingroup$


We have four coins




  • Coin 1: $0.10


  • Coin 2: $1.00


  • Coin 3: $1.00


  • Coin 4: $1.00



How many ways can we get $20.00 from these coins?



My attempt:



I started by counting the total number of ways for each coin to reach $20.00




  • 200 ways for coin 1


  • 20 ways for coins 2, 3, and 4.



I now have an equation, a + b + c + d + e = 200



We want to get the total number of solutions without any constraints



$dbinom{200+5-1}{5-1} = dbinom{204}{4}$



Then I found the number of solutions with the constraint that the coin must be $leq$ 20. The total number of solutions with coin > 20 can be determined by purchasing 21 coins and leaving 181 at most to buy.



$dbinom{181+5-1}{5-1} = dbinom{185}{4}$



The final solution is:



$$dbinom{204}{4} - 3 times dbinom{185}{4} $$






combinatorics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    If only four coins, why five vars a,b,c,d,e?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 18:20










  • $begingroup$
    'e' represents the difference between 200 and the number of coins purchased.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:21












  • $begingroup$
    I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 18:29










  • $begingroup$
    @saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:37
















2












$begingroup$


We have four coins




  • Coin 1: $0.10


  • Coin 2: $1.00


  • Coin 3: $1.00


  • Coin 4: $1.00



How many ways can we get $20.00 from these coins?



My attempt:



I started by counting the total number of ways for each coin to reach $20.00




  • 200 ways for coin 1


  • 20 ways for coins 2, 3, and 4.



I now have an equation, a + b + c + d + e = 200



We want to get the total number of solutions without any constraints



$dbinom{200+5-1}{5-1} = dbinom{204}{4}$



Then I found the number of solutions with the constraint that the coin must be $leq$ 20. The total number of solutions with coin > 20 can be determined by purchasing 21 coins and leaving 181 at most to buy.



$dbinom{181+5-1}{5-1} = dbinom{185}{4}$



The final solution is:



$$dbinom{204}{4} - 3 times dbinom{185}{4} $$






combinatorics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    If only four coins, why five vars a,b,c,d,e?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 18:20










  • $begingroup$
    'e' represents the difference between 200 and the number of coins purchased.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:21












  • $begingroup$
    I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 18:29










  • $begingroup$
    @saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:37














2












2








2


0



$begingroup$


We have four coins




  • Coin 1: $0.10


  • Coin 2: $1.00


  • Coin 3: $1.00


  • Coin 4: $1.00



How many ways can we get $20.00 from these coins?



My attempt:



I started by counting the total number of ways for each coin to reach $20.00




  • 200 ways for coin 1


  • 20 ways for coins 2, 3, and 4.



I now have an equation, a + b + c + d + e = 200



We want to get the total number of solutions without any constraints



$dbinom{200+5-1}{5-1} = dbinom{204}{4}$



Then I found the number of solutions with the constraint that the coin must be $leq$ 20. The total number of solutions with coin > 20 can be determined by purchasing 21 coins and leaving 181 at most to buy.



$dbinom{181+5-1}{5-1} = dbinom{185}{4}$



The final solution is:



$$dbinom{204}{4} - 3 times dbinom{185}{4} $$






combinatorics permutations







share|cite|improve this question











$endgroup$




We have four coins




  • Coin 1: $0.10


  • Coin 2: $1.00


  • Coin 3: $1.00


  • Coin 4: $1.00



How many ways can we get $20.00 from these coins?



My attempt:



I started by counting the total number of ways for each coin to reach $20.00




  • 200 ways for coin 1


  • 20 ways for coins 2, 3, and 4.



I now have an equation, a + b + c + d + e = 200



We want to get the total number of solutions without any constraints



$dbinom{200+5-1}{5-1} = dbinom{204}{4}$



Then I found the number of solutions with the constraint that the coin must be $leq$ 20. The total number of solutions with coin > 20 can be determined by purchasing 21 coins and leaving 181 at most to buy.



$dbinom{181+5-1}{5-1} = dbinom{185}{4}$



The final solution is:



$$dbinom{204}{4} - 3 times dbinom{185}{4} $$






combinatorics permutations




combinatorics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 18:27









Matt Samuel

38.8k63769




38.8k63769










asked Dec 6 '18 at 18:15









Arthur GreenArthur Green

796




796












  • $begingroup$
    If only four coins, why five vars a,b,c,d,e?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 18:20










  • $begingroup$
    'e' represents the difference between 200 and the number of coins purchased.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:21












  • $begingroup$
    I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 18:29










  • $begingroup$
    @saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:37


















  • $begingroup$
    If only four coins, why five vars a,b,c,d,e?
    $endgroup$
    – coffeemath
    Dec 6 '18 at 18:20










  • $begingroup$
    'e' represents the difference between 200 and the number of coins purchased.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:21












  • $begingroup$
    I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 18:29










  • $begingroup$
    @saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
    $endgroup$
    – Arthur Green
    Dec 6 '18 at 18:37
















$begingroup$
If only four coins, why five vars a,b,c,d,e?
$endgroup$
– coffeemath
Dec 6 '18 at 18:20




$begingroup$
If only four coins, why five vars a,b,c,d,e?
$endgroup$
– coffeemath
Dec 6 '18 at 18:20












$begingroup$
'e' represents the difference between 200 and the number of coins purchased.
$endgroup$
– Arthur Green
Dec 6 '18 at 18:21






$begingroup$
'e' represents the difference between 200 and the number of coins purchased.
$endgroup$
– Arthur Green
Dec 6 '18 at 18:21














$begingroup$
I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
$endgroup$
– saulspatz
Dec 6 '18 at 18:29




$begingroup$
I don't understand what you mean by $200$ ways for coin 1. If the twenty dollars consists solely of coins of type 1, there will be $200$ of them, but that's only one way, isn't it? In fact the number of coins of type 1 bust be a multiply of $10$, so there are only $21$ choices for the number of coins of type 1.
$endgroup$
– saulspatz
Dec 6 '18 at 18:29












$begingroup$
@saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
$endgroup$
– Arthur Green
Dec 6 '18 at 18:37




$begingroup$
@saulspatz I thought I could solve the problem using constraints. I wanted to take the total number of options, then remove the smaller constraints from that total.
$endgroup$
– Arthur Green
Dec 6 '18 at 18:37










1 Answer
1






active

oldest

votes


















4












$begingroup$

There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of $ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $binom{23}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 19:15










  • $begingroup$
    Thanks, @saulspatz.
    $endgroup$
    – Anubhab Ghosal
    Dec 6 '18 at 19:51











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of $ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $binom{23}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 19:15










  • $begingroup$
    Thanks, @saulspatz.
    $endgroup$
    – Anubhab Ghosal
    Dec 6 '18 at 19:51
















4












$begingroup$

There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of $ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $binom{23}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 19:15










  • $begingroup$
    Thanks, @saulspatz.
    $endgroup$
    – Anubhab Ghosal
    Dec 6 '18 at 19:51














4












4








4





$begingroup$

There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of $ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $binom{23}{3}$.






share|cite|improve this answer









$endgroup$



There should be a multiple of 10 number of coins of type $1$. Hence, in effect we have four types of coin of $ $1$ and desire non-negative solutions to the equation $x_1+x_2+x_3+x_4=20$, which by the stars and bars problem is $binom{23}{3}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 18:40









Anubhab GhosalAnubhab Ghosal

1,22319




1,22319












  • $begingroup$
    Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 19:15










  • $begingroup$
    Thanks, @saulspatz.
    $endgroup$
    – Anubhab Ghosal
    Dec 6 '18 at 19:51


















  • $begingroup$
    Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 19:15










  • $begingroup$
    Thanks, @saulspatz.
    $endgroup$
    – Anubhab Ghosal
    Dec 6 '18 at 19:51
















$begingroup$
Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
$endgroup$
– saulspatz
Dec 6 '18 at 19:15




$begingroup$
Very neat. I went through adding up the the stars and bars answer if there are $10k$ coins of type $1$, for $k=0,1,dots,20$ and eventually ended up with ${23choose3},$ but it never clicked that I could shortcut all that.
$endgroup$
– saulspatz
Dec 6 '18 at 19:15












$begingroup$
Thanks, @saulspatz.
$endgroup$
– Anubhab Ghosal
Dec 6 '18 at 19:51




$begingroup$
Thanks, @saulspatz.
$endgroup$
– Anubhab Ghosal
Dec 6 '18 at 19:51


















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