Solutions — Counting 10D Worksheet

Year 11 Mathematical Methods · Cambridge Methods 1&2, Counting Methods · Mr Wong

ANSWER KEY

Warm-up — set up the fraction

Q1. Letters of MATHS (5 distinct):

a $5!=120$ arrangements

b $\dfrac{4!}{5!}=\dfrac{1}{5}$ (fix S at the end)

c $\dfrac{4!}{5!}=\dfrac{1}{5}$ (fix M at the front)

Part A — Probability with arrangements

2a $\dfrac{2\times4!}{5!}=\dfrac{48}{120}=\dfrac{2}{5}$ (last digit 2 or 4)

2b $\dfrac{3\times4!}{5!}=\dfrac{72}{120}=\dfrac{3}{5}$ (first digit 3, 4 or 5)

3 $\dfrac{3!\times2!}{4!}=\dfrac{12}{24}=\dfrac{1}{2}$ (glue A, B)

4 $\dfrac{4!\times4!}{7!}=\dfrac{576}{5040}=\dfrac{4}{35}$ (boys as a block → 4 items, ×4! internal)

Part B — Probability with selections

Q5. Committee of $3$ from $5$ men, $3$ women; total $=\,^8C_3=56$:

a $^8C_3=56$ committees

b $\dfrac{^3C_3}{56}=\dfrac{1}{56}$ (all women)

c $\dfrac{^3C_2\,^5C_1}{56}=\dfrac{15}{56}$ (exactly 2 women)

d $1-\dfrac{^5C_3}{56}=1-\dfrac{10}{56}=\dfrac{23}{28}$ (at least 1 woman)

6 $\dfrac{^4C_3}{^{10}C_3}=\dfrac{4}{120}=\dfrac{1}{30}$ (all left-handed)

7 $\dfrac{^3C_2}{^7C_2}=\dfrac{3}{21}=\dfrac{1}{7}$ (both red)

Challenge

8a $\dfrac{^8C_3}{^9C_4}=\dfrac{56}{126}=\dfrac{4}{9}$

8b If Priya is in, the other $3$ members are chosen from the remaining $8$ students: $^8C_3=56$ favourable committees out of $^9C_4=126$ total, giving $\dfrac{4}{9}$.