Spm test question(Combination) [closed]
$begingroup$
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 '18 at 14:57
nafhgood
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
$endgroup$
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 '18 at 4:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 '18 at 14:57
nafhgood
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
$endgroup$
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 '18 at 4:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
1
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47
add a comment |
$begingroup$
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 '18 at 14:57
nafhgood
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
$endgroup$
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
combinatorics combinations factorial
edited Nov 27 '18 at 14:57
nafhgood
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
edited Nov 27 '18 at 14:57
nafhgood
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
edited Nov 27 '18 at 14:57
nafhgood
1,797422
edited Nov 27 '18 at 14:57
nafhgood
1,797422
edited Nov 27 '18 at 14:57
nafhgood
1,797422
1,797422
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
asked Nov 27 '18 at 14:30
lukhman rahimlukhman rahim
1
1
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 '18 at 4:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 '18 at 4:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
1
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47
add a comment |
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
1
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
1
1
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is important to note that
$${y choose m} = {y choose n} iff color{blue}{m+n = y}$$
Unless $m = n$, which is trivial. You cannot express $y$ in terms of $m$ and $n$ in that case because $y$ could be any positive integer.
This can be shown very easily.
$${y choose m} = {y choose n}$$
Substituting $color{blue}{m = y-n}$, you get
$${y choose color{blue}{y-n}} = {y choose n}$$
$$frac{y!}{(y-n)!(y-(y-n))!} = frac{y!}{n!(y-n)!}$$
$$frac{y!}{(y-n)!n!} = frac{y!}{n!(y-n)!}$$
Which is true. You can also prove this backwards.
$$frac{y!}{m!(y-m)!} = frac{y!}{(y-m)!(y-(y-m))!}$$
$${y choose m} = {y choose y-m}$$
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
$endgroup$
add a comment |
$begingroup$
There is an identity that
If $nchoose m$$=$$nchoose a$ then either $a=m$ or $a+m=n$ .
In your case $y=n+m$ .
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
$endgroup$
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is important to note that
$${y choose m} = {y choose n} iff color{blue}{m+n = y}$$
Unless $m = n$, which is trivial. You cannot express $y$ in terms of $m$ and $n$ in that case because $y$ could be any positive integer.
This can be shown very easily.
$${y choose m} = {y choose n}$$
Substituting $color{blue}{m = y-n}$, you get
$${y choose color{blue}{y-n}} = {y choose n}$$
$$frac{y!}{(y-n)!(y-(y-n))!} = frac{y!}{n!(y-n)!}$$
$$frac{y!}{(y-n)!n!} = frac{y!}{n!(y-n)!}$$
Which is true. You can also prove this backwards.
$$frac{y!}{m!(y-m)!} = frac{y!}{(y-m)!(y-(y-m))!}$$
$${y choose m} = {y choose y-m}$$
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
$endgroup$
add a comment |
$begingroup$
It is important to note that
$${y choose m} = {y choose n} iff color{blue}{m+n = y}$$
Unless $m = n$, which is trivial. You cannot express $y$ in terms of $m$ and $n$ in that case because $y$ could be any positive integer.
This can be shown very easily.
$${y choose m} = {y choose n}$$
Substituting $color{blue}{m = y-n}$, you get
$${y choose color{blue}{y-n}} = {y choose n}$$
$$frac{y!}{(y-n)!(y-(y-n))!} = frac{y!}{n!(y-n)!}$$
$$frac{y!}{(y-n)!n!} = frac{y!}{n!(y-n)!}$$
Which is true. You can also prove this backwards.
$$frac{y!}{m!(y-m)!} = frac{y!}{(y-m)!(y-(y-m))!}$$
$${y choose m} = {y choose y-m}$$
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
$endgroup$
add a comment |
$begingroup$
It is important to note that
$${y choose m} = {y choose n} iff color{blue}{m+n = y}$$
Unless $m = n$, which is trivial. You cannot express $y$ in terms of $m$ and $n$ in that case because $y$ could be any positive integer.
This can be shown very easily.
$${y choose m} = {y choose n}$$
Substituting $color{blue}{m = y-n}$, you get
$${y choose color{blue}{y-n}} = {y choose n}$$
$$frac{y!}{(y-n)!(y-(y-n))!} = frac{y!}{n!(y-n)!}$$
$$frac{y!}{(y-n)!n!} = frac{y!}{n!(y-n)!}$$
Which is true. You can also prove this backwards.
$$frac{y!}{m!(y-m)!} = frac{y!}{(y-m)!(y-(y-m))!}$$
$${y choose m} = {y choose y-m}$$
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
$endgroup$
It is important to note that
$${y choose m} = {y choose n} iff color{blue}{m+n = y}$$
Unless $m = n$, which is trivial. You cannot express $y$ in terms of $m$ and $n$ in that case because $y$ could be any positive integer.
This can be shown very easily.
$${y choose m} = {y choose n}$$
Substituting $color{blue}{m = y-n}$, you get
$${y choose color{blue}{y-n}} = {y choose n}$$
$$frac{y!}{(y-n)!(y-(y-n))!} = frac{y!}{n!(y-n)!}$$
$$frac{y!}{(y-n)!n!} = frac{y!}{n!(y-n)!}$$
Which is true. You can also prove this backwards.
$$frac{y!}{m!(y-m)!} = frac{y!}{(y-m)!(y-(y-m))!}$$
$${y choose m} = {y choose y-m}$$
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
edited Nov 27 '18 at 15:49
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
answered Nov 27 '18 at 15:12
KM101KM101
5,9381524
5,9381524
add a comment |
add a comment |
$begingroup$
There is an identity that
If $nchoose m$$=$$nchoose a$ then either $a=m$ or $a+m=n$ .
In your case $y=n+m$ .
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
$endgroup$
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
add a comment |
$begingroup$
There is an identity that
If $nchoose m$$=$$nchoose a$ then either $a=m$ or $a+m=n$ .
In your case $y=n+m$ .
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
$endgroup$
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
add a comment |
$begingroup$
There is an identity that
If $nchoose m$$=$$nchoose a$ then either $a=m$ or $a+m=n$ .
In your case $y=n+m$ .
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
$endgroup$
There is an identity that
If $nchoose m$$=$$nchoose a$ then either $a=m$ or $a+m=n$ .
In your case $y=n+m$ .
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:43
Akash RoyAkash Roy
1
1
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
add a comment |
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
$begingroup$
@lukhman rahim accept my answer
$endgroup$
– Akash Roy
Nov 28 '18 at 10:20
add a comment |
$begingroup$
Accept the answer if you have understood
$endgroup$
– Akash Roy
Nov 27 '18 at 14:43
1
$begingroup$
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
$endgroup$
– nafhgood
Nov 27 '18 at 14:47