##### I have no idea what to do???

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The answer is 41.

Explanation:

Case 1: The case where she makes 2 sales(S)

She cannot get any ‘fails’ (F),or the
log will end immediately.

She can also get from 0 to 4 tentatives
(T). (she does not have to get the full tentatives because she made 2 sales)

Let n(S), n(T), n(F) denote the number of sales, … etc.

So, when the log ends with n(S) = 2, n(F) = 0 and n(T) = 0, 1, …, 4

With n(T) = 0, there is just 1 possible log: S, S.

With n(T) = 1, there are 2 possible logs: T, S, S and S, T, S.

With n(T) = 2, the number of logs is the number of ways of choosing which position 1, 2 or 3 the S will come,=3

With n(T) = 3, the number of logs is 4.

With n(T) = 4, the number of logs is 5.

Add these totals to find the number of possible logs for case 1

Total for n(S) = 2 is 1 + 2 + 3 + 4 + 5 = 15

Thus case 1=15

Case 2

If n(T) = 5, then once again n(F) = 0, and n(S) = 0 or 1.

If n(S) = 0, there is just 1 log.

If n(S) = 1, there are 5 positions in which the S could come.

Add these totals to find the number of cases for case 2

So the total for n(T) = 5 is 1 + 5 = 6

Case 3

If n(F) = 1, then the F must come at the end of the sequence, with n(S) = 0, 1 and n(T) = 0, 1, …, 4

If n(S) = 0, there are 5 possible logs, corresponding to each of the five values of n(T).

If
n(S) = 1, there are 1 + 2 + 3 + 4 + 5 = 15 possible logs (each of these
individual numbers being the number of ways of positioning the S with
0, 1, …, 4 T’s)

Add these totals to find the number of case 3

So the total for n(F) = 1 is 5 + 15 = 20

Add all the totals from case 1, case 2, and case 3

Case 1+Case 2+Case 3= 15 + 6 + 20 = 41 logs.

**Received answer from http://askmemathsquestions.tumblr.com/page/36**

Edited some of it so it made a little more sense, but the math is sound.

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