2011 H2 Mathematics Paper 2 Question 11

Permutations and Combinations (P&C)

Answers

(i)

8168671≈0.0941{\frac{816}{8671}\approx0.0941}

(ii)

r=6{r=6}

Full solutions

(i)

P(R=4)=(184)(126)(3010)=8168671  ■\begin{align*} \textrm{P}\left(R=4\right) &= \frac{{18 \choose 4}{12 \choose 6}}{{30 \choose 10}} \\ &= \frac{816}{8671} \; \blacksquare \end{align*}

(ii)

P(R=r)>P(R=r+1)(18r)(1210−r)(3010)>(18r+1)(1210−(r+1))(3010)(18r)(1210−r)>(18r+1)(129−r)\begin{align*} \textrm{P}\left(R=r\right) &> \textrm{P}\left(R=r+1\right) \\ \frac{{18\choose r}{12 \choose 10-r}}{30 \choose 10} &> \frac{{18\choose r+1}{12 \choose 10-(r+1)}}{30 \choose 10} \\ {18\choose r}{12 \choose 10-r} &> {18\choose r+1}{12 \choose 9-r} \\ \end{align*}
18!r!(18−r)!⋅12!(10−r)!(r+2)!>18!(r+1)!(17−r)!⋅12!(9−r)!(r+3)!(r+1)!(17−r)!(9−r)!(r+3)!>r!(18−r)!(10−r)!(r+2)!  ■\begin{align*} \frac{18!}{r!(18-r)!}\cdot\frac{12!}{(10-r)!(r+2)!} &> \frac{18!}{(r+1)!(17-r)!}\cdot\frac{12!}{(9-r)!(r+3)!} \\ (r+1)!(17-r)!(9-r)!(r+3)! &> r!(18-r)!(10-r)!(r+2)! \; \blacksquare \end{align*}
(r+1)!r!(r+3)!(r+2)!>(18−r)!(17−r)!(10−r)!(9−r)!(r+1)(r+3)>(18−r)(10−r)r2+4r+3>180−28r+r2\begin{align*} \frac{(r+1)!}{r!} \frac{(r+3)!}{(r+2)!} &> \frac{(18-r)!}{(17-r)!}\frac{(10-r)!}{(9-r)!} \\ (r + 1)(r + 3) &> (18 - r)(10 - r) \\ r^2 + 4 r + 3 &> 180 - 28 r + r^2 \\ \end{align*}
32r>177r>17732r>5.5313\begin{align*} 32r &> 177 \\ r &> \frac{177}{32} \\ r &> 5.5313 \\ \end{align*}
Hence r=6  ■{r=6 \; \blacksquare}