Determine how many possible groups can be chosen.

Determine how many groups can be formed consisting of two males and two females.

Determine how many groups can be formed consisting of at least one female.

$\left( \begin{array}{l} 11\\ 4 \end{array} \right)$ $= \frac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}} = 330$     (M1)A1

[2 marks]

$\left( \begin{array}{l} 5\\ 2 \end{array} \right) \times \left( \begin{array}{l} 6\\ 2 \end{array} \right) = \frac{{5 \times 4}}{{2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}$     M1

$= 150$     A1

[2 marks]

METHOD 1

number of ways all men $=$ $\left( \begin{array}{l} 5\\ 4 \end{array} \right) = 5$

$330 - 5 = 325$     M1A1

Note:     Allow FT from answer obtained in part (a).

[2 marks]

METHOD 2

$\left( \begin{array}{l} 6\\ 1 \end{array} \right)\left( \begin{array}{l} 5\\ 3 \end{array} \right) + \left( \begin{array}{l} 6\\ 2 \end{array} \right)\left( \begin{array}{l} 5\\ 2 \end{array} \right) + \left( \begin{array}{l} 6\\ 3 \end{array} \right)\left( \begin{array}{l} 5\\ 1 \end{array} \right) + \left( \begin{array}{l} 6\\ 4 \end{array} \right)$     M1

$= 325$     A1

[2 marks]

Total [6 marks]