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There are n boys and m girls.A group of p people should be formed where there should be min 4 boys and 1 girl.Find total number of way of forming the group if number of boys n is 5,number of girls m=2 and total number of people p is 5.How to go about solving this problem

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Too special. If the committee with $p=5$ people must contain at least $4$ boys and at least $1$ girl, then it must have exactly $4$ boys and exactly $1$ girl. The $4$ boys can be chosen in $\binom{5}{4}=5$ ways, and for each choice, the girl can be chosen in $\binom{2}{1}=2$ ways. The product is $10$. – André Nicolas Feb 9 '12 at 17:30
ok dats using combination.Simply wat i asked.Just wanted to know if it can be solved by some other means. – Jay Feb 9 '12 at 17:37
Well, we don't need to use the symbol $\binom{n}{k}$. The $4$ boys can be chosen in $5$ ways (there are clearly $5$ ways to choose who won't be on the committee). And then there are $2$ ways to choose the girl. At a more basic level, we can list all the committees and count. Give the boys and girls names. Simplest way to list the committees is as $(X,Y)$, where $X$ is the boy who is left out, and $Y$ is the girl who is left out. – André Nicolas Feb 9 '12 at 17:44
up vote 2 down vote accepted

There are $5$ boys and $2$ girls and you want to form a group of $5$ with at least $4$ boys and $1$ girl. That means you should choose $1$ girl and $4$ boys. This can be done in $\binom 21\times \binom 54=10$ ways.

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