Spm test question(Combination) [closed]
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 at 14:57
mathnoob
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 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 |
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 at 14:57
mathnoob
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 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.
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
1
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47
add a comment |
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
edited Nov 27 at 14:57
mathnoob
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
$binom{y}{m}=binom{y}{n}$,
How should I express y in terms of m and n?
combinatorics combinations factorial
combinatorics combinations factorial
edited Nov 27 at 14:57
mathnoob
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
edited Nov 27 at 14:57
mathnoob
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
edited Nov 27 at 14:57
mathnoob
1,794422
edited Nov 27 at 14:57
mathnoob
1,794422
edited Nov 27 at 14:57
mathnoob
1,794422
1,794422
asked Nov 27 at 14:30
lukhman rahim
1
asked Nov 27 at 14:30
lukhman rahim
1
asked Nov 27 at 14:30
lukhman rahim
1
1
closed as off-topic by Adrian Keister, user10354138, Davide Giraudo, Brahadeesh, Aaron Montgomery Nov 28 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 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.
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
1
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47
add a comment |
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
1
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
1
1
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47
add a comment |
2 Answers
2
active
oldest
votes
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 at 15:12
KM101
4,836421
add a comment |
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 at 14:43
Akash Roy
1
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 15:12
KM101
4,836421
add a comment |
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 at 15:12
KM101
4,836421
add a comment |
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 at 15:12
KM101
4,836421
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 at 15:12
KM101
4,836421
edited Nov 27 at 15:49
answered Nov 27 at 15:12
KM101
4,836421
answered Nov 27 at 15:12
KM101
4,836421
answered Nov 27 at 15:12
KM101
4,836421
4,836421
add a comment |
add a comment |
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 at 14:43
Akash Roy
1
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
add a comment |
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 at 14:43
Akash Roy
1
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
add a comment |
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 at 14:43
Akash Roy
1
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 at 14:43
Akash Roy
1
answered Nov 27 at 14:43
Akash Roy
1
answered Nov 27 at 14:43
Akash Roy
1
answered Nov 27 at 14:43
Akash Roy
1
1
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
add a comment |
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
@lukhman rahim accept my answer
– Akash Roy
Nov 28 at 10:20
add a comment |
Accept the answer if you have understood
– Akash Roy
Nov 27 at 14:43
1
By definition $binom{n}{m}=frac{n!}{m!(n-m)!}=frac{n!}{(n-m)!(n-(n-m))!}=binom{n}{n-m}$.
– mathnoob
Nov 27 at 14:47