Answer:
96
Explanation:
COMBINATIONS & PERMUTATIONS can be confusing but they always have a solution.
Key 1: Understand and/or rewrite the question
Consider the lists of 6 items (out of the 5 letters P,R,O,F, and S; where O is used twice).
NOTE: When they say O is used more than once, don't forget that each list must not exceed 6 items in total. So, O is used twice. Simple!
How many such lists can be made if each list must end in an S?
NOTE: This instruction requires that S doesn't move. It won't change position, and it won't be involved in the rearrangement!
Key 2: Solve the mysteries
Now, S won't move and O is doubled. These rules put a form of restriction on the "n" which is the total number of items involved in the operation. We are supposed to create lists of 6 items but the reality is - only 5 are moving. Recall also that the original number of items involved in the operation is 5.
In this case, n = 5
Then we have 4 items rotating - P, R, O and F - with 1 repeated.
What this means is that for every list starting with the first O, the different arrangements apply to the second O as well.
There is no distinction such as O₁ and O₂. O is O!
Let k = the extra O
(n - k) = 5 - 1 = 4
So each of these 4 letters has 4! (4 factorial) arrangements, that is (4 x 3 x 2 x 1) = 24 arrangements
Multiply 24 by 4 to get 96