Prove, by combinatorial arguments, for $1\leq m\leq k\leq n$, this identity holds.
\[\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}\]
Identity
- Phlembac Adib Hasan
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- Fatin Farhan
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Re: Identity
Let $k$ be the number of all students, $n$ be the number of regional winners, $m$ be the number of national winners. We can choose regional winners from all participants in $\binom{n}{k}$ ways and national winners from regional winners in $\binom{k}{m}$ ways.
Again we can choose national winners from all participants in $\binom{n}{m}$ ways. As national winners are also regional winners, so rest of the regional winners can be chosen in $\binom{n-m}{k-m}$ ways.
So,$$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$$
Again we can choose national winners from all participants in $\binom{n}{m}$ ways. As national winners are also regional winners, so rest of the regional winners can be chosen in $\binom{n-m}{k-m}$ ways.
So,$$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$$
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- Phlembac Adib Hasan
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Re: Identity
$k\leq n$. উল্টাটা ধরতে হবে।Fatin Farhan wrote:Let $\color{red}{k}$ be the number of all students, $\color{red}{n}$ be the number of regional winners, $m$ be the number of national winners. We can choose regional winners from all participants in $\binom{n}{k}$ ways and national winners from regional winners in $\binom{k}{m}$ ways.
Again we can choose national winners from all participants in $\binom{n}{m}$ ways. As national winners are also regional winners, so rest of the regional winners can be chosen in $\binom{n-m}{k-m}$ ways.
So,$$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$$
In your proof, LHS=regional winners$\times $ national winners=national winners $\times $ rest of the regional winners=RHS. Why should they be equal?
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Re: Identity
In LHS, we first find the ways of selecting regional winners from a pool of participants, and then find the ways of selecting national winners from those regional winners (this also creates a class of regional winners who aren't national winners). So ultimately we have the number of ways to select several regional winners (who aren't national winners), and some national winners (who are also regional winners).Phlembac Adib Hasan wrote:$k\leq n$. উল্টাটা ধরতে হবে।Fatin Farhan wrote:Let $\color{red}{k}$ be the number of all students, $\color{red}{n}$ be the number of regional winners, $m$ be the number of national winners. We can choose regional winners from all participants in $\binom{n}{k}$ ways and national winners from regional winners in $\binom{k}{m}$ ways.
Again we can choose national winners from all participants in $\binom{n}{m}$ ways. As national winners are also regional winners, so rest of the regional winners can be chosen in $\binom{n-m}{k-m}$ ways.
So,$$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$$
In your proof, LHS=regional winners$\times $ national winners=national winners $\times $ rest of the regional winners=RHS. Why should they be equal?
In RHS, we find the ways of selecting national winners from participants, and then find the ways of selecting other regional winners (who are not national winners). Thus we conclude that LHS=RHS.
A bit redundant perhaps, but I think it's rigorous enough.