2010 H2 Mathematics Paper 2 Question 8

Permutations and Combinations (P&C)

Answers

(i)

35{\frac{3}{5}}

(ii)

110{\frac{1}{10}}

(iii)

720{\frac{7}{20}}

Full solutions

(i)

P(number greater than 30,000)=3×4!5!=35  ■\begin{align*} & \textrm{P}\left(\textrm{number greater than } 30,000\right) \\ &= \frac{3 \times 4!}{5!} \\ &= \frac{3}{5} \; \blacksquare \end{align*}

(ii)

P(last two digits both even)=3!×2!5!=110  ■\begin{align*} & \textrm{P}\left(\textrm{last two digits both even}\right) \\ &= \frac{3! \times 2!}{5!} \\ &= \frac{1}{10} \; \blacksquare \end{align*}

(iii)

Case 1: first digit is 3{3} or 5{5}. Probability
=(21)×(21)×3!5!=15\begin{align*} & = \frac{{2 \choose 1} \times {2 \choose 1} \times 3!}{5!} \\ & = \frac{1}{5} \end{align*}
Case 2: first digit is 4{4}. Probability
=(11)×(31)×3!5!=320\begin{align*} & = \frac{{1 \choose 1} \times {3 \choose 1} \times 3!}{5!} \\ & = \frac{3}{20} \end{align*}
Required probability=15+320=720  ■\begin{align*} & \textrm{Required probability} \\ & = \frac{1}{5} + \frac{3}{20} \\ & = \frac{7}{20} \; \blacksquare \end{align*}