Permutations and combinations

Basic solution:

This is like filling $5$ places with $5$ no.s so it should be equivalent to $5! = 120$

However, it is critical to note that the letter R is repeated twice which means that $IR_{1}OPR$ is different to $IR_{2}OPR$ which is not what we want.

Since there are $2 R's$ we will $\dfrac{5!}{2!} = \dfrac{120}{2} = \large \color{#3D99F6} \boxed{60}$

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Check out the wiki on Permutations

If $P$ represents the number of distinct permutations of $n$ things taken all at a time when, of the $n$ things, there are $u$ alike, $v$ others alike, $w$ others alike, etc., then

$P=\dfrac{n!}{u!v!w!...}$

In the word $PRIOR$ , there are two $R's$ , so

$P=\dfrac{5!}{2!}=\boxed{60}$