Question

There are 7 children in a classroom. In how many ways can they line up for recess?

Answer 1

Hint:This obvious problem is a permutation. The difference between permutations and combinations, recall, is that order matters in permutations. This is a permutation, considering that the question asks how many ways the students should line up for recess (i.e. in how many different orders). Use permutation to solve this problem.

Complete step by step solution:
To solve this problem, we have a General formula for determining the number of permutations of “n” taken items “r”. We tend to use the formula at a time without substitution (i.e., the student in position 1 will not move to the reception area and it becomes an alternative for position 2),
Number of permutation $ = \dfrac{{n!}}{{(n - r)!}}$
With “n” is the numbers of objects, and “r” is the number of vacancies which should be filled. If we compare this formula to the question, then we have $7$ number of students to be filled or stand in $7$ number of vacancies.
That means, $n = 7\;{\text{and}}\;r = 7$
Applying the permutation formula, we will get
Number of ways $ = \dfrac{{7!}}{{(7 - 7)!}} = \dfrac{{7!}}{{0!}} = \dfrac{{7!}}{1} = 5040$
Therefore only $7$ children can stand in a row or line up for recess in $5040$ ways.

Note: We take the value of $0! = 1.$ This problem can be solved without the knowledge of permutation or its formula by imagining the number of ways to be filled for the first place then second place then third up to the last place. It can be understood as there are seven available choices for first place but six for the second place (because first has been taken by one) similarly for the last one only one choice.