2008 H2 Mathematics Paper 2 Question 10

Permutations and Combinations (P&C)

Answers

(i)

120{120}

(ii)

9{9}

(iii)

210{210}

(iv)

485{485}

Full solutions

(i)

Number of ways
=(32)×(43)×(53)=120  ■\begin{align*} &= {3 \choose 2} \times {4 \choose 3} \times {5 \choose 3} \\ &= 120 \; \blacksquare \end{align*}

(ii)

Number of ways
=(98)=9  ■\begin{align*} &= {9 \choose 8} \\ &= 9 \; \blacksquare \end{align*}

(iii)

Case 1: Exactly 4 diplomats from M.{M.} Number of ways
=(54)×(74)=175\begin{align*} & = {5 \choose 4} \times {7 \choose 4} \\ & = 175 \end{align*}
Case 2: Exactly 5 diplomats from M.{M.} Number of ways
=(55)×(73)=35\begin{align*} & = {5 \choose 5} \times {7 \choose 3} \\ & = 35 \end{align*}
Required umber of ways
=175+35=210  ■\begin{align*} & = 175 + 35 \\ & = 210 \; \blacksquare \end{align*}

(iv)

We will consider the complement, which consists of two cases
Complement case 1: No diplomat from K{K}
Number of ways
=(98)=9\begin{align*} & = {9 \choose 8} \\ & = 9 \end{align*}
Complement case 2: No diplomat from L{L}
Number of ways
=(88)=1\begin{align*} & = {8 \choose 8} \\ & = 1 \end{align*}
Required number of ways
=(128)−9−1=485  ■\begin{align*} & = {12 \choose 8} - 9 - 1 \\ & = 485 \; \blacksquare \end{align*}