Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote -1 down vote favorite

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?

share|improve this question
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

1 Answer

up vote -1 down vote

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 $

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.