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]

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

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