Solution — 2023 Question 7 Permutations & Combinations

(a)

For the sum to be odd, we need an odd number of odd numbers in our selection. From ${1, \dots, 5}$ there are 3 odd and 2 even numbers.

Case 1: 1 odd, 2 even.
Number of ways $= (\binom{3}{1}) (\binom{2}{2})$

Case 2: 3 odd, 0 even.
Number of ways $= (\binom{3}{3}) (\binom{2}{0})$

\begin{matrix} P (sum is odd) \\ = \frac{(\binom{3}{1}) (\binom{2}{2}) + (\binom{3}{3}) (\binom{2}{0})}{(\binom{5}{3})} \\ = \frac{2}{5} ∎ \end{matrix}

(b)

We will use the complement, since it’s easier to calculate if they have no common numbers.

$Total ways = (\binom{9}{2}) \times (\binom{9}{2})$

For no common numbers, after Aran picks two numbers from nine, Ben must choose two from the remaining 7 to have no common numbers.

$n (No common) = (\binom{9}{2}) \times (\binom{7}{2})$

\begin{matrix} P (at least one in common) \\ = 1 - P (no common numbers) \\ = 1 - \frac{(\binom{9}{2}) \times (\binom{7}{2})}{(\binom{9}{2}) \times (\binom{9}{2})} \\ = \frac{5}{12} ∎ \end{matrix}