Specimen 2017 H2 Mathematics Paper 2 Question 5

Permutations and Combinations (P&C)

Answers

(i)

6!×(2!)6=46,080.{6! \times (2!)^6 = 46,080.}

(ii)

2,678,400.{2,678,400.}

(iii)

140.{140.}

Full solutions

(i)

Required number of ways=6!×(2!)6=46,080  ■\begin{align*} & \textrm{Required number of ways} \\ & = 6! \times (2!)^6 \\ & = 46,080 \; \blacksquare \end{align*}

(ii)

Case 1: Wives at one corner
Number of ways=6!×(51)×5!×2=864,000\begin{align*} & \textrm{Number of ways} \\ & = 6! \times {5 \choose 1} \times 5! \times 2 \\ &= 864,000 \end{align*}
Case 2: Husband of wife on right end sits on left side of wives
Number of ways=6!×(51)×5!=432,000\begin{align*} & \textrm{Number of ways} \\ & = 6! \times {5 \choose 1} \times 5! \\ &= 432,000 \end{align*}
Case 3: Husband of wives in the middle sit on left side of wives
Number of ways=6!×(41)×(41)×5!=1,382,400\begin{align*} & \textrm{Number of ways} \\ & = 6! \times {4 \choose 1} \times {4 \choose 1} \times 5! \\ &= 1,382,400 \end{align*}
Required number of ways=864,000+432,000+1,382,400=2,678,400  ■\begin{align*} & \textrm{Required number of ways} \\ &= 864,000 + 432,000 + 1,382,400 \\ & = 2,678,400 \; \blacksquare \end{align*}

(iii)

Case 1: Exactly 1 wife
Number of ways=(61)×(52)=60\begin{align*} & \textrm{Number of ways} \\ & = {6 \choose 1} \times {5 \choose 2} \\ &= 60 \end{align*}
Case 2: Exactly 2 wives
Number of ways=(62)×(41)=60\begin{align*} & \textrm{Number of ways} \\ & = {6 \choose 2} \times {4 \choose 1} \\ &= 60 \end{align*}
Case 3: 3 wives
Number of ways=(63)=20\begin{align*} & \textrm{Number of ways} \\ & = {6 \choose 3} \\ &= 20 \end{align*}
Required number of ways=60+60+20=140  ■\begin{align*} & \textrm{Required number of ways} \\ &= 60 + 60 + 20 \\ & = 140 \; \blacksquare \end{align*}