Solutions — Counting 10B Worksheet

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

ANSWER KEY

Warm-up — factorials and nPr

Q1. Evaluations:

a $24$

b $720$

c $5040$

d $^7P_2=42$

e $^8P_3=336$

f $^{12}P_4=11\,880$

Q2. Boxes $4\times3\times2\times1=$ $24$ orders.

Part A — Arranging in a row

3a $7!=5040$

3b $^7P_3=7\times6\times5=210$

4 $^9P_4=9\times8\times7\times6=3024$

5 $6!=720$ arrangements of NUMBER

Part B — Restrictions: even & end positions

6a $2\times4!=48$ even (last digit 2 or 4)

6b $3\times4!=72$ (first digit 3, 4 or 5)

Part C — Items together or apart

7a $5!=120$

7b $4!\times2!=48$ together (glue AB, ×2! internal)

7c $120-48=72$ not together

8 $7!\times2!=10\,080$ (pair as a block of 8→7 items)

9a $9!\times2!=725\,760$ girls together

9b $10!-9!\times2!=2\,903\,040$ not together

Challenge

10a $7200$ arrangements

10b Place $C$ first: $2$ end seats. With $C$ fixed, arrange the other $7$ in $7!$ ways, then subtract those with $A,B$ adjacent (treat $AB$ as a block among the $7$: $6!\times2!$). So one end gives $7!-6!\times2!=5040-1440=3600$; times $2$ ends $=$ $7200$. Dealing with $C$ first, then using total $-$ (AB together), avoids double-counting.