Home / CIE A level Math -Probability & Statistics 1: 5.2 Permutations and combinations: objects in a line : Exam Style Questions Paper 5

# CIE A level Math -Probability & Statistics 1: 5.2 Permutations and combinations: objects in a line : Exam Style Questions Paper 5

### Question

(a) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be
arranged so that all 3 Es are together.                                                                                                             [2]

(b) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be
arranged so that the Ps are not next to each other.                                                                                      [4]

(c) Find the probability that a randomly chosen arrangement of the 10 letters of the word
SHOPKEEPER has an E at the beginning and an E at the end.                                                                 [2]

Four letters are selected from the 10 letters of the word SHOPKEEPER.
(d) Find the number of different selections if the four letters include exactly one P.                                 [3]

Ans

7 (a) $$\frac{8!}{2!}$$

20160

7 (b) Total number of ways: $$\frac{10!}{2!3!}(=302400)(A)$$

With Ps together: $$\frac{9!}{3!}(=60480)(B)$$

With Ps not together: 302 400 – 60 480

241 920
Alternative method for question 7(b)

$$\frac{8!}{3!}$$

$$\times \frac{9\times 8}{2}$$

241 920

7 (c) $$Probability=\frac{Number\ of\ ways\ Es\ at\ beginning\ and\ end}{Total\ number\ of\ ways}$$

$$Probability=\frac{\frac{8!}{2!}}{\frac{10!}{2!\times 3!}}=\frac{20160}{302400}$$

$$\frac{1}{15}, 0·0667$$

Alternative method for question 7(c)

$$Probability =\frac{3}{10}\times \frac{2}{9}$$

$$\frac{1}{15}, 0·0667$$

Alternative method for question 7(c)

$$Probability=\frac{1}{10}\times \frac{1}{9}\times 3!$$

$$\frac{1}{15},0.0667$$

7(d) Scenarios:
P E E E  5C0 = 1
P E E _  5C1 = 5
P E _ _  5C2 = 10
P _ _ _  5C3 = 10

Total = 26

### Question

(a) Find the number of different arrangements that can be made from the 9 letters of the word
JEWELLERY in which the three Es are together and the two Ls are together.                                [2]

(b) Find the number of different arrangements that can be made from the 9 letters of the word
JEWELLERY in which the two Ls are not next to each other.                                                              [4]

Ans

2( a) 6!
720

2 (b) Total number: $$\frac{9!}{3!2!}(30240)$$

Number with Ls together $$\frac{8!}{3!}(6720)$$

Number with Ls not together $$\frac{9!}{3!2!}-\frac{8!}{3!}$$

= 30 240 – 6720
23 520

Alternative method for question 2(b)

$$\frac{7!}{3!}\times \frac{8\times 7}{2}$$

7! × k in numerator, k integer ≥ 1

8 × 7 × m in numerator or 8C2 × m, m integer ≥ 1
3! in denominator
23 520

### Question

(a) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR.
(b) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR in
which there is an R at the beginning and an R at the end, and the two As are not together.
(c) Find the total number of different selections of 6 letters from the 11 letters of the word
CATERPILLAR that contain both Rs and at least one A and at least one L.

Ans:

1. $$\frac{11!}{2!2!2!}$$
4989600

2. Method 1 R^^^^^^^R
Arrange the 7 letters CTEPILL =$$\frac{7!}{2!}$$
Number of ways of placing As in non-adjacent places = $$^8C_2$$
$$\frac{7!}{2!}\times ^8C_2$$
=70560
Method 2 [Arrangements Rs at ends- Arrangements Rs at ends and As together]
Total arrangements with R at beg. and end $$=\frac{9!}{2!2!}$$
Arrangements with R at ends and As together $$=\frac{8!}{2!}$$
With As not together $$=\frac{9!}{2!2!}$$
With As not together $$=\frac{9!}{2!2!}-\frac{8!}{2!}$$
[90720-20160]=70560

3. Method 1
RRAL__ $$^5C_2=10$$
RRALL_ $$^5C_1=5$$
RRAAL_ $$^5C_1=5$$
RRAALL =1
[Total =]21
Method 2 – Fixing RRAL first.
N.B No other scenarios can be present anywhere in solution.
RRAL^^=$$^5C_2$$
[Total=]21

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