Question:
In how many ways can 5 children stand in a queue?
Solution:
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.