Number of permutations of 5 boys sitting in a row in such a way that at least one boy sticks to his original seat

Question

Question is attached in the fig.

My approach: Total no. of permutations of 5 boys = 5! = 120

No. of required arrangements = Total no. of permutations - no. of permutations in which no boy sits in his original seat

Now ,how can i find no. of permutations in which no boy sits in his original seat? or how can i solve this problem??

Answer 1

Your approach looks sound.

A permutation where no element stays put is known as a derangement.

An easy formula for the number of derangements of $n$ elements is $\frac{n!}{e}$ rounded to the nearest integer. So for your exercise you're looking for the integer nearest to $(1-\frac1e)\cdot 5!$

(In an exercise where you had to provide proof of your claim something as simple as this wouldn't do, of course -- then you'd have to trace out the proof of the count of the number of derangements -- but for multiple choice all tricks count).

Answer 2

You're going right. For finding no. of permutations in which no boy sits in his original seat use dearrangement?

In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position. In other words, derangement is a permutation that has no fixed points.