Consider the lists of length six made with the symbols P, R, O, F, S, where repetition is allowed. (For example, the following is such a list: (P, R,O, O,F, S).) How many such lists can be made if the list must end in an S and the symbol O is used more than once?

96

Step-by-step 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

answer: the answer is 40.82

step-by-step explanation: 3.14 x 13(or half of 26) to the power of 2

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step-by-step explanation: