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Probability Statements on Letter Arrangements of SYLLABUS

CMAT 2023 · Quantitative Techniques & Data Interpretation · Medium · Probability

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A number of distinct 8-letter words are possible using the letters of the word SYLLABUS. If a word is chosen at random, then Statement I: The probability that the word contains the two S's together is 14\frac{1}{4}. Statement II: The probability that the word begins and ends with L is 128\frac{1}{28}. In the light of the above statements, choose the correct answer from the options given below:

  1. A.

    Both Statement I and Statement II are true

  2. B.

    Both Statement I and Statement II are false

  3. C.

    Statement I is true but Statement II is false

  4. D.

    Statement I is false but Statement II is true

Answer

A

Explanation

Letters: S, Y, L, L, A, B, U, S (8 letters: 2 S's, 2 L's). Total words = 8!2!2!\frac{8!}{2!2!}.

Statement I: Treat (SS) as 1 unit. Words = 7!2!⇒Prob=7!/2!8!/(2!2!)=28=14\frac{7!}{2!} \Rightarrow \text{Prob} = \frac{7! / 2!}{8! / (2!2!)} = \frac{2}{8} = \frac{1}{4}. (True)

Statement II: Fix L at start and end. Remaining 6 letters (S, Y, A, B, U, S with 2 S's) arranged in 6!2!\frac{6!}{2!} ways. Prob=6!/2!8!/(2!2!)=28×7=128\text{Prob} = \frac{6! / 2!}{8! / (2!2!)} = \frac{2}{8 \times 7} = \frac{1}{28}. (True)

Both statements are true.

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