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In a group of $x$ people, there should be a minimum of $n$ boys and minimum of $m$ girls. Need to find the number of ways the group can be formed, where $4 \leq n \leq 100$, $1 \leq m \leq 20$ and $5 \leq x \leq n+m$.

What is the simplest way to find the solution?

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What sort of pool are you selecting from? If it's all boys or all girls, the answer becomes easy. –  ncmathsadist Feb 9 '12 at 15:22
    
The problem is incompletely specified. As mentioned by ncmathsadist, we need to know how many boys and how many girls we are making our selection of $x$ people from. If we are assuming indistinguishability (so we are producing a collection of $x$ red and blue balls), the answer is easy. Assume $x\ge m+n$. We have freedom on how to select the remaining $x-(m+n)$. Of these, $0$ to $x-(m+n)$ can be girls, for a total of $x-(m+n)+1$ ways. –  André Nicolas Feb 9 '12 at 15:31
    
updated the condition –  Jay Feb 9 '12 at 16:05
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1 Answer 1

Let's denote :

$N_g$ - number of girls and $N_b$ - number of boys ,

$k$ - number of ways that group can be formed . Then :

$N_b \geq n$ and $ N_g \geq m$ , so :

$N_g=m \Rightarrow N_b = x -m$

$N_g=m+1 \Rightarrow N_b = x -(m+1)$

$\vdots$

$N_g=m+k' \Rightarrow N_b=x-(m+k')$

Therefore :

$x-(m+k')=n \Rightarrow k'=x-(m+n) \Rightarrow $

$\Rightarrow k=x-(m+n)+1$

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