We are given a class consisting of 4 boys and 4 girls. a committee that consists of a President, a Vice-President and a secretary is to be chosen among the 8 students of the class. Let a denote the number of ways of choosing the committee in such a way that the committee has at least one boy and atleast one one girl. Let b denote the number of ways of choosing the committee in such a way that the number of girls is greater than or equal to that of boys. Then what are the values of a and b?
[Math] We are given a class consisting of 4 boys and 4 girls.
combinatoricspermutations
Best Answer
As Simon said in his answer above, $a$ can have two cases:
$a.1:$ 2 boys 1 girl, which yields $\binom{4}{2}\binom{4}{1}$ choices for people and $3!$ arrangements for their positions which yields $6\cdot4\cdot6=144$ possibilities.
$a.2:$ 1 boy 2 girls, which is the same as above and yields $144$.
Thus $a=288$.
$b$ also has two cases:
$b.1:$ 1 boy 2 girls, which as above, yields $144$.
$b.2:$ 3 girls, which yields $\binom{4}{3}$ to choose the three girls and $3!$ arrangements which gives $4\cdot6$ possibilities $24$.
Thus $b=168$.