2021 H2 Mathematics Paper 2 Question 7

Permutations and Combinations (P&C)

Answers

(i)

83,160{83,160}

(ii)

144{144}

(iii)

166{\frac{1}{66}}

Full solutions

(i)

Number of different arrangements
=11!5!  2!  2!=83,160  ■\begin{align*} &= \frac{11!}{5!\;2!\;2!} \\ &= 83,160 \; \blacksquare \end{align*}

(ii)

∘ B B ∘ R R × A A A A × C D ∘\circ \, \boxed{\textrm{B B}} \, \circ \, \boxed{\textrm{R R}} \, \times \, \boxed{\textrm{A A A A}} \, \times \, \boxed{\textrm{C D}} \, \circ
Number of different arrangements
=4!×2!×(31)=144  ■\begin{align*} &= 4! \times 2! \times {3 \choose 1} \\ &= 144 \; \blacksquare \end{align*}

(iii)

B B R R C D A A A A A\textrm{B B R R C D } \boxed{\textrm{A A A A A}}
P(5A’s together)=7!2!  2!83160=166  ■\begin{align*} & \textrm{P}\left(5 \textrm{A's together}\right) \\ & = \frac{\frac{7!}{2!\;2!}}{83160} \\ & = \frac{1}{66} \; \blacksquare \end{align*}