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For all positive integers n,m,k where $n\ge m\ge k$ , $\binom {n}{m}\binom {m}{k}=\binom {n}{k}\binom {n-k}{n-m}$

Prove the following statements using combinatorial proofs.

I can't come up combinatorial proofs.

Also,

$\sum_{i=0}^n {n \choose i} = 2^n$

how would you proof $2^n$?

puppylord
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  • It is slightly more natural to do it in two steps. First argue that $\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$ and then doing the easy combinatorial argument that $\binom{n-k}{m-k}=\binom{n-k}{n-m}$. – André Nicolas Nov 15 '13 at 07:52
  • @André Nicolas Do you think that is a typo? or it is possible to proof $\binom {n-k}{n-m}$=$\binom {n-k}{m-k}$ – puppylord Nov 15 '13 at 15:06
  • @puppylord It is not a typo. Note that $\binom {n}{k}=\binom {n}{n-k}$ – hhsaffar Nov 15 '13 at 15:31

4 Answers4

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Select $m$ out of $n$ club members to become board mebers and then select $k$ out of the board mebers to become board member spokespersons. Or select $k$ board meber spokespersons and then select $n-m$ out of the remaining $n-k$ club members who shall not join the board. Since both methods are equivalent, equality holds. (A bit more formally we count the number of pairs $(A,B)$ where $A\subseteq B\subseteq\{1,\ldots,n\}$ and $|A|=k$ and $|B|=m$ in two ways)

In a set of $n$ elements, there are $n\choose i$ subsets of size $i$. If we sum over all $i$ we get the totoal number of subsets, which is $2^n$.

Hagen von Eitzen
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1

Out of n members here we have to count in how many ways we can choose a committee of m members and out of those m members we have to choose a sub-committee of k members ... clearly the right hand sight denotes the possible number of ways we can do this thing!.... Another way to count the same thing is to count that how many ways we can form our sub-committee which is nothing but $\binom {n}{k}$ and then we choose the other members of the committee and surely this can be done in $\binom {n-k}{m-k}$ ways so int the second case we form the the same committee by $\binom {n}{k}\binom {n-k}{m-k}$ ways claiming the fact!

And for your $2^n$ proof the LHS denotes the total number of subsets of a n-set and another way to count the same thing is for any particular member, say the $\alpha$-th member can be selected in our subset or not so it has 2 possibility with it and this thing is same for the other member.... So all together we can select subsets in $2^n$ ways, done!

tensor
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Given $n$ people, we can form a committee of size $m$ in ${n\choose m}$ ways. Once the committee is formed we can form a sub-committee of size $k$ in ${m\choose k}$ ways. Thus we can form a committee of size $m$ with a sub-committee of size $k$ in ${n\choose m}{m\choose k}$ 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 $k$ in ${n\choose k}$ ways. Once the sub-committee is formed we must form the committee of size $n-m$ from the remaining $n-k$ people in ${n-k\choose n-m}$ ways. Thus we can form a sub-committee of size $k$ while forming the committee of size $n-m$ that contains the sub-committee in ${n\choose k}{n-k\choose n-m}$ ways. Hence ${n\choose m}{m\choose k}={n\choose k}{n-k\choose n-m}$.

This combinatorial identity is known as the Subset-of-a-Subset identity.

Now, we must show that $${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}=2^n$$ is the number of subsets of an $n$-element set $S$ where $n\geq0$.

Every subset of $S$ is an $i$-subset of $S$ where $i=0,1,2,...,n$. We know that ${n\choose i}$ equals the number of $i$-subsets of S. Thus by the Addition Principle $${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}$$ equals the number of subsets to the set $S$. We can count the same thing by observing that each element of the set $S$ has two choices, either they are in a subset or they are not in a subset. Let $S=\{x_1,x_2,x_3,...,x_n\}$. So, $x_1$ is either in a subset or it is not in a subset, $x_2$ is either in a subset or it is not in a subset,..., $x_n$ is either in a subset or it is not in a subset. Thus by the Multiplication Principle there are $2^n$ ways we can form a subset of the set $S$. Hence ${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}=2^n$.

Another approach is to consider the Binomial Theorem $$(x+y)^n=\sum_{i=0}^n {n\choose i}x^{n-i}y^k.$$ Letting $x=1$ and $y=1$ we obtain$$2^n=\sum_{i=0}^n{n\choose i}.$$

1233dfv
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  • wow, very clear..But if we want to write a combinatorial proofs for second question, am I suppose to use committee instead of subset? – puppylord Nov 15 '13 at 14:32
  • You can write your combinatorial proofs in any way you want. The second proof is a standard proof of that identity. It can proved in a much faster way. However, when doing these proofs it is important to be thorough so that the concepts are understood well. – 1233dfv Nov 15 '13 at 14:53
  • Also, my question is $\binom {n-k}{n-m}$ not $\binom {n-k}{m-k}$, is that a typo or it is same? $\binom {n-k}{n-m}$=$\binom {n-k}{m-k}$? – puppylord Nov 15 '13 at 15:03
  • They're the same. It is the symmetry rule for the binomial coefficients. – 1233dfv Nov 15 '13 at 17:32
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Hint: For the first one:Suppose there are $n$ items and you want to take $m$ of them with yourself. You have two bags, one bag can hold $k$ items and another can take $m-k$ items, in how many ways can you take the items with your self?

For the second one:A set has $n$ members, the number of subsets with $0$ members $+$ the number of subsets with 1 member $+$ the number of subsets with 2 members $+$...$+$ the number of subsets with n members is?

hhsaffar
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