Prove $binomnabinomn-ab-a = binomnbbinomba$ The 2019 Stack Overflow Developer Survey Results Are InShow that $beginalignn choose kk choose m = n choose mn-m choose k-m.endalign$For all positive integers n,m,k where $nge mge k$ , $binom nmbinom mk=binom nkbinom n-kn-m$Prove $binom kmbinom tk=binom tmbinomt-mt-k$Combinatorial Proof that $binomnvphantomdcbinomn-db-c=binomnvphantomddbinomdc$Stuck on combinatorial proof that $binomwp binompm = binomwm binomw-mp-m$The Hexagonal Property of Pascal's TriangleStrehl identity for the sum of cubes of binomial coefficientsShow $binomnkbinomka = binomnabinomn-ak-a$ by block-walking interpretation of Pascal's triangleProve $binomnm binommk = binomnk binomn-km-k$ by counting in two waysTrying to find $sumlimits_k=0^n k binomnk$How to prove Vandermonde's Identity: $sum_k=0^nbinomRkbinomMn-k=binomR+Mn$?Fermat's Combinatorial Identity: How to prove combinatorially?How to prove that $sum_k=0^n binom nk k^2=2^n-2(n^2+n)$Elementary combinatorial questionProve $kbinom nk=n binomk-1 n-1$ algebraically.Combinatorics Question Help; # of ways to choose 4 distinct officials from a city?Proving $binom n-1r-1=sum_k=0^r(-1)^kbinom r k binomn+r-k-1r-k-1$Prove that $sumlimits_k=0^mbinommkbinomnr+k=binomm+nm+r$ using combinatoric arguments.Prove using combinatorics $sumlimits_i=k^n-r+kbinomikbinomn-ir-k=binomn+1r+1$
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Prove $binomnabinomn-ab-a = binomnbbinomba$
The 2019 Stack Overflow Developer Survey Results Are InShow that $beginalignn choose kk choose m = n choose mn-m choose k-m.endalign$For all positive integers n,m,k where $nge mge k$ , $binom nmbinom mk=binom nkbinom n-kn-m$Prove $binom kmbinom tk=binom tmbinomt-mt-k$Combinatorial Proof that $binomnvphantomdcbinomn-db-c=binomnvphantomddbinomdc$Stuck on combinatorial proof that $binomwp binompm = binomwm binomw-mp-m$The Hexagonal Property of Pascal's TriangleStrehl identity for the sum of cubes of binomial coefficientsShow $binomnkbinomka = binomnabinomn-ak-a$ by block-walking interpretation of Pascal's triangleProve $binomnm binommk = binomnk binomn-km-k$ by counting in two waysTrying to find $sumlimits_k=0^n k binomnk$How to prove Vandermonde's Identity: $sum_k=0^nbinomRkbinomMn-k=binomR+Mn$?Fermat's Combinatorial Identity: How to prove combinatorially?How to prove that $sum_k=0^n binom nk k^2=2^n-2(n^2+n)$Elementary combinatorial questionProve $kbinom nk=n binomk-1 n-1$ algebraically.Combinatorics Question Help; # of ways to choose 4 distinct officials from a city?Proving $binom n-1r-1=sum_k=0^r(-1)^kbinom r k binomn+r-k-1r-k-1$Prove that $sumlimits_k=0^mbinommkbinomnr+k=binomm+nm+r$ using combinatoric arguments.Prove using combinatorics $sumlimits_i=k^n-r+kbinomikbinomn-ir-k=binomn+1r+1$
$begingroup$
I want to prove this equation,
$$
binomnabinomn-ab-a = binomnbbinomba
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
combinatorics discrete-mathematics binomial-coefficients
$endgroup$
add a comment |
$begingroup$
I want to prove this equation,
$$
binomnabinomn-ab-a = binomnbbinomba
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
combinatorics discrete-mathematics binomial-coefficients
$endgroup$
add a comment |
$begingroup$
I want to prove this equation,
$$
binomnabinomn-ab-a = binomnbbinomba
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
combinatorics discrete-mathematics binomial-coefficients
$endgroup$
I want to prove this equation,
$$
binomnabinomn-ab-a = binomnbbinomba
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
combinatorics discrete-mathematics binomial-coefficients
combinatorics discrete-mathematics binomial-coefficients
edited Nov 17 '13 at 9:00
Arjang
5,65962364
5,65962364
asked Oct 21 '13 at 6:44
More CodeMore Code
1464
1464
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
$endgroup$
add a comment |
$begingroup$
Directly, we find that
beginalign*
n choose an - a choose b - a &= fracn!(n - a)!a! frac(n - a)!(n - a - (b - a))! (b - a)! \
&= fracn!a! frac1(n - b)!(b - a)! \
&= fracn!(n - b)! b! fracb!(b - a)!a! \
&= n choose bb choose a
endalign*
The third equality follows by multiplying and dividing by $b!$.
$endgroup$
add a comment |
$begingroup$
These are just two different ways to express the trinomial coefficient $binom na,b-a,n-b$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
binom nk,l,m = binom nkbinom n-kl qquadtextwhenever $k+l+m=n$,
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $binom nb=binom nn-b$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
$endgroup$
add a comment |
$begingroup$
Given $n$ people, we can form a committee of size $b$ in $nchoose b$ ways. Once the committee is formed we can form a sub-committee of size $a$ in $bchoose a$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in $nchoose bbchoose a$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in $nchoose a$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in $n-achoose b-a$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in $nchoose an-achoose b-a$ ways. Hence $nchoose bbchoose a=nchoose an-achoose b-a$.
This combinatorial identity is known as the Subset-of-a-Subset identity.
$endgroup$
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
add a comment |
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4 Answers
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active
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votes
4 Answers
4
active
oldest
votes
active
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active
oldest
votes
$begingroup$
First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
$endgroup$
add a comment |
$begingroup$
First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
$endgroup$
add a comment |
$begingroup$
First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
$endgroup$
First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
edited Oct 23 '13 at 19:38
answered Oct 21 '13 at 6:50
André NicolasAndré Nicolas
455k36432821
455k36432821
add a comment |
add a comment |
$begingroup$
Directly, we find that
beginalign*
n choose an - a choose b - a &= fracn!(n - a)!a! frac(n - a)!(n - a - (b - a))! (b - a)! \
&= fracn!a! frac1(n - b)!(b - a)! \
&= fracn!(n - b)! b! fracb!(b - a)!a! \
&= n choose bb choose a
endalign*
The third equality follows by multiplying and dividing by $b!$.
$endgroup$
add a comment |
$begingroup$
Directly, we find that
beginalign*
n choose an - a choose b - a &= fracn!(n - a)!a! frac(n - a)!(n - a - (b - a))! (b - a)! \
&= fracn!a! frac1(n - b)!(b - a)! \
&= fracn!(n - b)! b! fracb!(b - a)!a! \
&= n choose bb choose a
endalign*
The third equality follows by multiplying and dividing by $b!$.
$endgroup$
add a comment |
$begingroup$
Directly, we find that
beginalign*
n choose an - a choose b - a &= fracn!(n - a)!a! frac(n - a)!(n - a - (b - a))! (b - a)! \
&= fracn!a! frac1(n - b)!(b - a)! \
&= fracn!(n - b)! b! fracb!(b - a)!a! \
&= n choose bb choose a
endalign*
The third equality follows by multiplying and dividing by $b!$.
$endgroup$
Directly, we find that
beginalign*
n choose an - a choose b - a &= fracn!(n - a)!a! frac(n - a)!(n - a - (b - a))! (b - a)! \
&= fracn!a! frac1(n - b)!(b - a)! \
&= fracn!(n - b)! b! fracb!(b - a)!a! \
&= n choose bb choose a
endalign*
The third equality follows by multiplying and dividing by $b!$.
answered Oct 21 '13 at 6:47
user61527
add a comment |
add a comment |
$begingroup$
These are just two different ways to express the trinomial coefficient $binom na,b-a,n-b$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
binom nk,l,m = binom nkbinom n-kl qquadtextwhenever $k+l+m=n$,
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $binom nb=binom nn-b$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
$endgroup$
add a comment |
$begingroup$
These are just two different ways to express the trinomial coefficient $binom na,b-a,n-b$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
binom nk,l,m = binom nkbinom n-kl qquadtextwhenever $k+l+m=n$,
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $binom nb=binom nn-b$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
$endgroup$
add a comment |
$begingroup$
These are just two different ways to express the trinomial coefficient $binom na,b-a,n-b$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
binom nk,l,m = binom nkbinom n-kl qquadtextwhenever $k+l+m=n$,
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $binom nb=binom nn-b$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
$endgroup$
These are just two different ways to express the trinomial coefficient $binom na,b-a,n-b$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
binom nk,l,m = binom nkbinom n-kl qquadtextwhenever $k+l+m=n$,
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $binom nb=binom nn-b$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
answered Oct 23 '13 at 12:17
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
88.7k5111230
add a comment |
add a comment |
$begingroup$
Given $n$ people, we can form a committee of size $b$ in $nchoose b$ ways. Once the committee is formed we can form a sub-committee of size $a$ in $bchoose a$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in $nchoose bbchoose a$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in $nchoose a$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in $n-achoose b-a$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in $nchoose an-achoose b-a$ ways. Hence $nchoose bbchoose a=nchoose an-achoose b-a$.
This combinatorial identity is known as the Subset-of-a-Subset identity.
$endgroup$
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
add a comment |
$begingroup$
Given $n$ people, we can form a committee of size $b$ in $nchoose b$ ways. Once the committee is formed we can form a sub-committee of size $a$ in $bchoose a$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in $nchoose bbchoose a$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in $nchoose a$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in $n-achoose b-a$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in $nchoose an-achoose b-a$ ways. Hence $nchoose bbchoose a=nchoose an-achoose b-a$.
This combinatorial identity is known as the Subset-of-a-Subset identity.
$endgroup$
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
add a comment |
$begingroup$
Given $n$ people, we can form a committee of size $b$ in $nchoose b$ ways. Once the committee is formed we can form a sub-committee of size $a$ in $bchoose a$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in $nchoose bbchoose a$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in $nchoose a$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in $n-achoose b-a$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in $nchoose an-achoose b-a$ ways. Hence $nchoose bbchoose a=nchoose an-achoose b-a$.
This combinatorial identity is known as the Subset-of-a-Subset identity.
$endgroup$
Given $n$ people, we can form a committee of size $b$ in $nchoose b$ ways. Once the committee is formed we can form a sub-committee of size $a$ in $bchoose a$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in $nchoose bbchoose a$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in $nchoose a$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in $n-achoose b-a$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in $nchoose an-achoose b-a$ ways. Hence $nchoose bbchoose a=nchoose an-achoose b-a$.
This combinatorial identity is known as the Subset-of-a-Subset identity.
edited Oct 25 '13 at 18:09
answered Oct 23 '13 at 1:15
1233dfv1233dfv
4,2031326
4,2031326
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
add a comment |
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
1
1
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Not serious objection: the two procedures are not equivalent, since in the second solution, the soldiers go to war untrained! (Also if you use $a$ as a variable, please put dollars around it.)
$endgroup$
– Marc van Leeuwen
Oct 23 '13 at 12:04
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
$begingroup$
Nice catch! I'll edit it to the more familiar committee argument. Thank you.
$endgroup$
– 1233dfv
Oct 25 '13 at 18:12
add a comment |
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