2014 H2 Mathematics Paper 2 Question 6

Permutations and Combinations (P&C)

Answers

(i)

31,500{31,500}

(ii)

16,800{16,800}

(iii)

8,820{8,820}

Full solutions

(i)

Number of different teams
=(31)×(84)×(52)×(64)=31,500  ■\begin{align*} & = {3 \choose 1} \times {8 \choose 4} \times {5 \choose 2} \times {6 \choose 4} \\ & = 31,500 \; \blacksquare \end{align*}

(ii)

Case 1: include the midfielder brother
Number of different teams
=(31)×(84)×(41)×(54)=4,200\begin{align*} & = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 1} \times {5 \choose 4} \\ & = 4,200 \end{align*}
Case 2: include the attacker brother
Number of different teams
=(31)×(84)×(42)×(53)=12,600\begin{align*} & = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 2} \times {5 \choose 3} \\ & = 12,600 \end{align*}
Total number of different teams
=4200+12600=16,800  ■\begin{align*} & = 4200 + 12600 \\ & = 16,800 \; \blacksquare \end{align*}

(iii)

Case 1: the special midfielder plays as a midfielder
Number of different teams
=(31)×(84)×(31)×(54)=3,150\begin{align*} & = {3 \choose 1} \times {8 \choose 4} \times {3 \choose 1} \times {5 \choose 4} \\ & = 3,150 \end{align*}
Case 2: the special midfielder plays as a defender
Number of different teams
=(31)×(83)×(32)×(54)=2,520\begin{align*} & = {3 \choose 1} \times {8 \choose 3} \times {3 \choose 2} \times {5 \choose 4} \\ & = 2,520 \end{align*}
Case 3: the special midfielder does not play
Number of different teams
=(31)×(84)×(32)×(54)=3,150\begin{align*} & = {3 \choose 1} \times {8 \choose 4} \times {3 \choose 2} \times {5 \choose 4} \\ & = 3,150 \end{align*}
Total number of different teams
=3150+2520+3150=8,820  ■\begin{align*} & = 3150 + 2520 + 3150 \\ & = 8,820 \; \blacksquare \end{align*}