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It's subsumed under Vandermonde's Identity.

Let a set consist of $n$ kiwis and $n$ figs. We want to count how many ways there are to select $n$ fruits, regardless of type. One immediate answer is $\binom{2n}{n}$.

Now count in a different way. Each selection of $n$ fruits contains a certain number, say $k$, of kiwis and also $n-k$ figs, so $0 \leq k \leq n$.
For any particular value of $k$, we can choose the kiwis in $\binom{n}{k}$ ways and the figs in $\binom{n}{n - k}$ ways.

So the # of ways to choose an $n$ set with $k$ kiwis is $\binom{n}{k}\binom{n}{n - k} = \binom{n}{k}^2$.

  1. I initially asked this here, but now separately.

Thus, $\color{green}{\text{the total # of ways to select $n$ fruits independent of type}}$
= total # of ways to select $0$ kiwis + total # of ways to select $1$ kiwis $+ \cdots +$ total # of ways to select $(n - 1)$ kiwis + $\color{orangered}{\text{the total # of ways to select $n$ KIWIS}}$ $\color{green}{= \binom{n}{0}^2 + ... \binom{n}{n - 1}^2 + \color{orangered}{\binom{n}{n}^2}}$. $\Large{\color{#B22222}{]}}$

  1. Why must we sum from $0 \leq k \leq n$?

3 Answers3

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If you have chosen $k$ kiwis, then you need to choose $n-k$ figs. The number of ways of choosing $n-k$ figs from $n$ figs is ${n \choose n-k}$ which is equal to ${n \choose k}$. Multiply this by the number of ways of choosing $k$ kiwis from $n$ and you get the number of ways of choosing $k$ kiwis and $n-k$ figs.

To get all possibilities, you have to consider all possibilities, ranging from choosing no kiwis to choosing all kiwis, i.e. choosing from $k=0$ to $k=n$ kiwis, so you need to sum over this range.

Henry
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  • Thank you. I'm still befogged so could you please respond to my supplementary in my OP, in your answer and NOT as a comment (to lodge your writings in one place)? –  Nov 17 '13 at 05:40
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  1. No, picking exactly $k$ kiwis means you discount the $n-k$ remaining kiwis, but you still have to chose $n-k$ figs out of $n$, which is the second factor.
  2. You can pick anywhere from $0$ (none) to $n$ (all) kiwis. $n+1$ is not allowed ($=0$ possibilities) and "adding" a kiwi (choosing $-1$ ~ adding a $n+1$-st Kiwi to the stack) is also disallowed.
AlexR
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  • Thank you. I'm still befogged so could you please respond to my supplementary in my OP, in your answer and NOT as a comment (to lodge your writings in one place)? –  Nov 17 '13 at 05:39
  • Also, could you please explicate what you mean by "adding" a kiwi $(−1)$? –  Nov 17 '13 at 05:39
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Since there are ${2n\choose n}$ ways to select $n$ pieces of fruit regardless of the type we can select $k$ of one type of fruit and the remaining $n-k$ of the other type of fruit. We then sum for all possible $k$ values from $0\leq k\leq n$ to obtain every possible combination.

1233dfv
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