Paper 2 Section B Practice: Permutations and Combinations — Answers

Questions: Permutations and Combinations Drills.

Question 1

Choose the first letter in $2$ ways, then arrange two of the remaining five letters in $5P_2$ ways. Choose the first digit from $1,2,3,4$ and the second from the remaining four digits. Hence

$$2(5P_2)(4)(4)=640.$$

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Question 2

(a)

$$\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{7}{3}\right)=210.$$

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(b) Subtract the all-student committees from all committees:

$$\left(\genfrac{}{}{0ex}{}{11}{5}\right)-\left(\genfrac{}{}{0ex}{}{7}{5}\right)=462-21=441.$$

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(c) Start from the 210 committees in part (a). Those containing both Jia and Noor require one further student from the other five and two teachers:

$$\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{5}{1}\right)=30.$$

Therefore the required number is

$$210-30=180.$$

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Question 3

The total number of distinct arrangements is

$$\frac{10!}{3!3!2!}=50400.$$

If the three $S$‘s form one block, the eight visible objects are the $SSS$ block, three $T$‘s, two $I$‘s, $A$ and $C$. Their number of arrangements is

$$\frac{8!}{3!2!}=3360.$$

Hence the required number is

$$50400-3360=47040.$$

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Question 4

(a) Treat Ali and Bea as a block. There are six circular objects and two internal orders:

$$2(6-1)!=240.$$

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(b)

$$(7-1)!-240=720-240=480.$$

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(c) Treat the consecutive triple as one block. Chen must be in the middle, while Ali and Bea can exchange sides. There are five circular objects, so

$$2(5-1)!=48.$$

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