In how many ways can 5 children stand in a queue?

To find: number of arrangements of 5 children in a queue.

Here, 5 places are needed to be occupied by 5 children.

Therefore any one of the 5 children can occupy first place.

Similarly, any 4 children can occupy second place and so on.

Lastly, there will be a single person to occupy the 5 position

Formula:

Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is

$P(n, r)=n ! /(n-r) !$

Therefore, permutation of 5 different objects in 5 places is

$P(5,5)=\frac{5 !}{(5-5) !}$

$=\frac{5 !}{0 !}=\frac{120}{1}=120$

Hence, this can be done in 120 ways.