GATE CSE 1990 | Question: 3-ix

Question

The number of ways in which $5\; A's, 5\; B's$ and $5\; C's$ can be arranged in a row is:

• A. $15!/(5!)^{3}$
• B. $15!$
• C. $\left(\frac{15}{5}\right)$
• D. $15!(5!3!)$.

Comments

why there’s a multi-select tag?

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Option A

15! / 5!.5!.5!

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15C5 *1 *10C5 *1 *5C5 *1  = 15! /(5!)^3

Answer 1

(A) Use permutation with repetitions formula as we have to arrange $15$ elements where $5$ each are identical. 

Comments

Total 15 and each one is repeating 5 times

So answer will 15!/5!5!5! or 15!/(5!)^3

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@Lakshman Bhaiyaunderstood sir

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It will be solved  by permutation with limited repetition 

because two indications is given in this question 

first: 5a's 5b's and 5c's - it means all a's are identical and all b's are identical and all c's are identical

second: keyword arrange is given in the question and it indicates permutation

so permutation with limited repetition is applied here or mississippi problem will be applied here

answer = 15!/5!5!5! =>15!/(5!)^3

Answer 2

The longer way

Option  $\large A$

Answer 3

MISSISSIPPI Problem

So, answer = 15!/(5!)³

Answer 4

A