2015 H2 Mathematics Paper 2 Question 11

Permutations and Combinations (P&C)

Answers

(i)

10,080{10,080}

(ii)

10,079{10,079}

(iii)

720{720}

(iv)

5,760{5,760}

Full solutions

(i)

Number of different arrangements
=82!2!=10,080  ■\begin{align*} & = \frac{8}{2!2!} \\ & = 10,080 \; \blacksquare \end{align*}

(ii)

Number of different arrangements =10080−1=10,079  ■{=10080-1} \allowbreak {= 10,079 \; \blacksquare}

(iii)

Number of different arrangements =6!=720  ■{=6! = 720 \; \blacksquare}

(iv)

Let A{A} denote the set of events where the two A{\textmd{A}}s are together and B{B} denote the set of events where the two B{\textmd{B}}s are together
Then (A∪B)′{\left(A\cup B \right)'} represents the required sets of arrangements
n(A)=n(B)=72!=2520\begin{align*} n(A) &= n(B) \\ &= \frac{7}{2!} \\ &= 2520 \end{align*}
n(A∪B)=n(A)+n(B)−n(A∩B)=2520+2520−720=4,320\begin{align*} n\left(A\cup B\right) &= n(A) + n(B) - n(A\cap B) \\ &= 2520 + 2520 - 720 \\ &= 4,320 \end{align*}
Require number of different arrangements
=n((A∪B)′)=10080−4320=5,760  ■\begin{align*} &= n\Big((A\cup B)'\Big) \\ &= 10080 - 4320 \\ &= 5,760 \; \blacksquare \end{align*}