Probability an interviewer gets her interviews

Question

If an interviewer needs to interview 5 people, and each person has an independent probability of $2/3$ of accepting her invitation, what is the probability that she speaks to exactly 6 people given that she has a list of 8 people?
I though that this was answerable by the binomial distribution, namely $p=C(8,6)(2/3)^6(1/3)^2$ however, the book says the answer is $160/729$
How did the book get to that result?

Answer 1

Note that by the way the answer is posed we understand we care not about which people we talked to. We only care about how many people we contacted.

So if she had to invite 6 people to be able to perform 5 interviews, it is because she invited someone who declined. Imagine you have an ordered list of people. The interviewer had to invite the first 6 people of that list in order to give 5 interviews.

Can you get the right answer now?

If $A $ means interview accepted and $D $ means interview declined then we want to count how many sequences of the type

$$AADAAA$$ there are, with the $D $ only being possible on some of the spots.

Answer 2

I figured out (RSerrao too). She will stop contacting people if (1) she have interviewed 5 people or if (2) she reached the end of the list.

Anyway for this problem, length of the list isn't matter, I guess it's given to confuse you. In the first 5 calls she needs to have exactly one decline and at 6th call she needs to be accepted. ${5\choose 1}\frac 1 3 (\frac 2 3)^4\frac 2 3=160/729$