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]
Answer/Explanation
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]
Answer/Explanation
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