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Permutations of 10-Digit Number

CMAT 2021 · Quantitative Techniques & Data Interpretation · Medium · Permutations and Combinations

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How many ten-digit numbers can be formed using all the digits of 2435753228 such that odd digits appear only in even places?

  1. A.

    2!3!5!2! 3! 5!

  2. B.

    (5!)2(5!)^2

  3. C.

    (5!)23!\frac{(5!)^2}{3!}

  4. D.

    (5!)23!(2!)2\frac{(5!)^2}{3!(2!)^2}

Answer

D

Explanation

The given number 2435753228 has 10 digits: 5 odd digits (3, 3, 5, 5, 7) and 5 even digits (2, 2, 2, 4, 8). The odd digits must go into the 5 even places, which can be arranged in 5!2!2!\frac{5!}{2! 2!} ways. The even digits go into the remaining 5 odd places, which can be arranged in 5!3!\frac{5!}{3!} ways. Total number of arrangements = 5!2!2!×5!3!=(5!)23!(2!)2\frac{5!}{2! 2!} \times \frac{5!}{3!} = \frac{(5!)^2}{3!(2!)^2}.

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