Distribution problem # 2 2

Probability Level pending

In how many ways can the following 11 11 letters : A A , B B , C C , D D , E E , F F , X X , X X , X X , Y Y , Y Y be arranged in a row so that every Y Y lies between two X ′ s X's ? ( not necessarily adjacent )


The answer is 997920.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

At first consider the X's and Y's only. These five letters (3X and 2Y) can be arranged in 5!÷(2!×3!) ways. Out of these 3 are the desired arrangements. Therefore the probability of getting a desired arrangement is 3/10. Now, the eleven letters can be arranged in 11!÷(3!×2!)=3326400 ways. 3/10 parts of these total permutations contain the desired arrangements of 3X and 2Y. So required numbers of permutations are (3/10)×3326400=997920.

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