Number of ways to arrange COLORADO with the last letter being O Obviously without the last letter restriction there are 8!/3! number of permutations. Clearly, we remove one of the Os and get 7! permutations (if the Os were distinguishable). Now, when accounting for the "double counting" do we divide by 2! or 3!? Clearly we could've  set aside any of the three Os and so I would think we still need to do 7!/3!
 A: Well, the remaining letters can indeed be permutated in $7! $ ways, but then for each permutation we can also interchange the two O's, so we have to divide by $2!=2 $. The final answer is hence $7!/2!=2520 $.
A: Let us count the ways to arrange COLPRADQ with the last letter being either O, P, or Q.
Now there are $3$ ways to select which to put in last place, and there are $7!$ ways to arrange the remaining letters.
Therefore there are $\;3\times 7!\;$ ways to arrange COLPRADQ with the last letter being either O, P, or Q.
Now if we replace the P and Q with Os, so that they become indistinct from the other, then every arrangement is now one of $3!$ equivalent arrangements.
So the count of distinct arrangements of COLORADO where the last letter is one of the Os, is $\dfrac{3\times 7!}{3!}$.
This is $\;\dfrac{7!}{2!}\;$ due to cancellation.

This is the same result as counting ways to:  Select any one of the three O, then arrange the remaining seven symbols including the remaining two indistinguishable Os.  That is: $1\times \frac{7!}{2!}$

Does that help you intuit the solution?
