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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?

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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 $b=168$.

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Case a is either 2 boys 1 girl or 1 boy 2 girls, case b is either 1 boy 2 girls or 3 girls. The numbers are small enough to enumerate the possibilities.

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  • $\begingroup$ A mathematical answer would help, regardless. If OP encounters a larger data set, it might not be 'small enough' and the equations would make more sense. $\endgroup$ – JTP - Apologise to Monica Jun 15 '15 at 10:07
  • $\begingroup$ You're right, but since there's only 8 possibilities in this case I wouldn't take the trouble. $\endgroup$ – Simon Jun 15 '15 at 10:09
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I agree with the answer by Linus S and the comments by Simon and Joe Taxpayer. This is an alternate approach using a generating function that will give the solutions handily even if the numbers are not "small enough".

We are counting the number of injective functions from the set {President, Vice President, Secretary} into the set {boy1, boy2, boy3, boy4, girl1, girl2, girl3, girl4}. The bivariate exponential generating function:

(1 + x)^4 * (1 + y*x)^4 will count the number of girls that are selected for an arbitrary number of positions. In this case (since there are 3 positions) we seek the coefficient of x^3/3! which is: 24 + 144 y + 144 y^2 + 24 y^3. This tells us that of all 336 possible selections there are 24 with no girls, 144 with exactly one girl, 144 with exactly 2 girls, and 24 with exactly 3 girls.

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