18N.1.AHL.TZ0.H_2

pestleMathematicsAAHLPaper 118N· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

A team of four is to be chosen from a group of four boys and four girls.

Find the number of different possible teams that could be chosen.

[3]

Find the number of different possible teams that could be chosen, given that the team must include at least one girl and at least one boy.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\left( {\begin{array}{*{20}{c}} 8 \\ 4 \end{array}} \right)$ (A1)

$= \frac{{8{\text{!}}}}{{4{\text{!}}4{\text{!}}}} = \frac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1}} = 7 \times 2 \times 5$ (M1)

= 70 A1

METHOD 2

recognition that they need to count the teams with 0 boys, 1 boy… 4 boys M1

$1 + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + 1$

$= 1 + \left( {4 \times 4} \right) + \left( {6 \times 6} \right) + \left( {4 \times 4} \right) + 1$ (A1)

= 70 A1

[3 marks]

EITHER

recognition that the answer is the total number of teams minus the number of teams with all girls or all boys (M1)

70 − 2

recognition that the answer is the total of the number of teams with 1 boy,

2 boys, 3 boys (M1)

$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) = \left( {4 \times 4} \right) + \left( {6 \times 6} \right) + \left( {4 \times 4} \right)$

THEN

= 68 A1

[2 marks]

Examiners’ report

[N/A]