20N.2.AHL.TZ0.H_7

pestleMathematicsAAHLPaper 220N· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

At a gathering of $12$ teachers, seven are male and five are female. A group of five of these teachers go out for a meal together. Determine the possible number of groups in each of the following situations:

There are more males than females in the group.

[4]

Two of the teachers, Gary and Gerwyn, refuse to go out for a meal together.

[3]

Markscheme / solution

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

identifying two or three possible cases (M1)

total number of possible groups is $(\begin{matrix} 7 \\ 5 \end{matrix}) + (\begin{matrix} 7 \\ 4 \end{matrix}) (\begin{matrix} 5 \\ 1 \end{matrix}) + (\begin{matrix} 7 \\ 3 \end{matrix}) (\begin{matrix} 5 \\ 2 \end{matrix})$ (A1)(A1)

Note: Award A1 for any two correct cases, A1 for the other one.

$= 21 + (35 \times 5) + (35 \times 10)$

$= 546$ A1

[4 marks]

METHOD 1

identifying at least two of the three possible cases- Gary goes, Gerwyn goes or neither goes (M1)

total number of possible groups is $(\begin{matrix} 10 \\ 5 \end{matrix}) + (\begin{matrix} 10 \\ 4 \end{matrix}) + (\begin{matrix} 10 \\ 4 \end{matrix})$ (A1)

$= 252 + 210 + 210$

$= 672$ A1

METHOD 2

identifying the overall number of groups and no. of cases where both Gary and Gerwyn go. (M1)

total number of possible groups is $(\begin{matrix} 12 \\ 5 \end{matrix}) - (\begin{matrix} 10 \\ 3 \end{matrix})$ (A1)

$= 792 - 120$

$= 672$ A1

[3 marks]

Examiners’ report

[N/A]