A problem on Arrangements by Nikola Alfredi

Probability Level 3

In how many ways you can arrange 20 boys and 5 girls sitting in form of a circle, such that there are atleast 2 boys between each girl.

Note : Atleast any 2 boys between any 2 girls.

The sum of the digits of the number is N (where N is total arrangements). Then N = n 2 − 1 N = n^2 - 1 . What is n n ?


The answer is 10.

Nikola Alfredi by Nikola Alfredi , 18, India

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1 solution

Muflih Maxi
Feb 28, 2020

Because they form a circle, let the position of one of the girl fixed. The number of ways to choose the 2nd, 3rd, 4th, and 5th girl is 4! (factorial). Let the number of boys between the girls substracted by 2 are a 1 a_1 , a 2 a_2 , a 3 a_3 , a 4 a_4 , a 5 a_5 , then a 1 + a 2 + a 3 + a 4 + a 5 = 10 a_1+a_2+a_3+a_4+a_5=10 . There are ( 14 4 ) {14}\choose{4} ways. And because the boys are distinguishable, there are 20! ways. So, the number of ways to do that is 4 ! ∗ 20 ! ∗ 4!*20!* ( 14 4 ) {14}\choose{4} = 58448037844435599360000 58448037844435599360000 . The sum of the digits are 99, therefore n=10 \fbox{n=10}

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Yes, I did the same way.

Nikola Alfredi - 1 year, 3 months ago

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