Question

3. A family has 5 children. If the family has 3 boys, how many different birth gender orders are possible? (20 points)

Answer 1

$\\ a! = a(a-1)(a-2)..1 \\ 0! = 1 \\\\ $ Number of ways of choosing "r" items from total "n" items $ \\\\ = \binom{n}{r} = \dfrac{n!}{r!(n-r)!} \\\\ ---------------------------- \\\\ $ The family has 5 children : n = 5 $ \\ $ Number of boys $ (r) = 3 \\\\ $ Number of different birth gender orders $ \\\\ = $ Number of ways of selecting 3 children out of 5 that will be boys $ \\ ($ Rest 2 will be automatically girls $) \\\\ = \binom{5}{3} = \dfrac{5!}{3!(5-3)!} \\\\ =\dfrac{5 * 4 * \cancel{3!}}{2 * 1 * \cancel{3!}} = \textbf{10}$