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Question: Five persons A, B, C, D and E occupy the seats in a row such that A and B sit next to each other. In...

Five persons A, B, C, D and E occupy the seats in a row such that A and B sit next to each other. In how many possible ways can these five people sit?

Explanation

Solution

In order to solve the given question , we should know the three important concepts related to the question: Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. We can say that permutation relates to the act of arranging all the members of a set into some sequence or order. And about combination , we can say it is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Here we are going to use the formula of permutation to get the required solution.

Complete step by step solution:
We know that if we have nn different objects to arrange , then :
The total number of arrangements = n!=n(n−1)(n−2)(n−3)....×3×2×1n! = n(n - 1)(n - 2)(n - 3).... \times 3 \times 2 \times 1
Now in our question it is given that there are 5 different persons which means 5 people can sit in 5!5! ways .
⇒5×4×3×2×1=120\Rightarrow 5 \times 4 \times 3 \times 2 \times 1 = 120 ways
Now , we are given that the 2 persons A and B must sit next to each other .
So, we will consider A and B as a single entity or as a single person
And now the total arrangements would be 4!4! ⇒4×3×2×1=24 \Rightarrow 4 \times 3 \times 2 \times 1 = 24, that is A and B can be arranged in 2!2! while seated together ( i.e. AB or BA ) 2!=2×1=22! = 2 \times 1 = 2
The total different ways in which 5 people can sit are 4!×2!=24×2=484! \times 2! = 24 \times 2 = 48
Therefore , The total different ways in which 5 people can sit are 4848.

Note:
If we are performing one operation AND another operation to carry out a single task when the outcome of one operation does not affect the number of possible outcomes of the other operation , then we MULTIPLY the number of possible outcomes .
Formula for permutation: nPr=n!(n−r)!=n(n−1)(n−2)......(n−r+1){}_{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} = n(n - 1)(n - 2)......(n - r + 1)