N boys and M girls are learning acting skills from a theatre in Mumbai. To perform a play on ‘Ramayana’ they need to form a group of P actors containing not less than 4 boys and not less than 1 girl. The theatre requires you to write a program that tells them the number of ways the group can be formed.
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Assume that $M \ge 1$, $N\ge 4$, and $P\ge 5$. In how many ways can we choose $P$ people, with no sex restrictions? We are choosing $P$ people from $M+N$. The number of choices is $$\tbinom{M+N}{P}.$$ However, choosing $0$, $1$, $2$, or $3$ boys is forbidden. In how many ways can we choose $0$ boys, and therefore $P$ girls? The answer is $\binom{M}{P}$. For reasons that will be apparent soon, we write that as $\binom{N}{0}\binom{M}{P}$. There are $\binom{N}{1}\binom{M}{P-1}$ ways to choose $1$ boy and $P-1\,$ girls. There are $\binom{N}{2}\binom{M}{P-2}$ ways to choose $2$ boys and $P-2\,$ girls. There are $\binom{N}{3}\binom{M}{P-3}$ ways to choose $3$ boys and $P-3\,$ girls. We need to deal with a small technical complication. Suppose, for example, that we are choosing $0$ boys and therefore $P$ girls, but the number $M$ of available girls is less than $P$. The answer above remains correct if we use the common convention that $\binom{m}{n}=0$ if $m<n$. Finally, at least $1$ girl must be chosen. So the $\binom{N}{P}\binom{M}{0}$ all-boy choices are forbidden. The total number of forbidden choices is therefore $$\tbinom{N}{0}\tbinom{M}{P}+\tbinom{N}{1}\tbinom{M}{P-1}+\tbinom{N}{2}\tbinom{M}{P-2} +\tbinom{N}{3}\tbinom{M}{P-3}+\tbinom{N}{P}\tbinom{M}{0}.$$ Finally, subtract the above number from the total number $\binom{M+N}{P}$ of choices with no sex restrictions. |
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