combinatorics: ways to answer 8 test Q's correctly, etc

a) On a 10-question test, how many ways are there to answer exactly eight questions correctly?

For this one, we figured it was just C(10, 8). Is this correct?

b) Repeat part (a) with the requirement that the first or second question, but not both, are answered correctly

We guessed on this one, but have no confidence in it.

c) Repeat part (a) in the case that three of the first five questions are answered correctly.

? <-- if I could, this would be a lot bigger

a) Yes.

b) 1st one correct and 2nd one wrong leaves 8 questions remaining. Then C(8,7) will be the total number of ways of answering 8 questions correctly overall. Repeating this for 1st one wrong and 2nd one correct gives another C(8,7) ways. Then total will be 2*C(8,7).

c) Choose 3 from the first 5 questions and 5 from the remaining 5 to get 8 correct overall. C(5,3)*C(5,5)=?

Part (a) is correc t.

For part (b), \(\displaystyle 2 {8 \choose 6}\)

For part (c) \(\displaystyle {5 \choose 3}\)

Thank you.

Hello, b_as_a_constant!

a) On a 10-question test, how many ways are there to answer exactly eight questions correctly?

For this one, we figured it was just C(10, 8). Is this correct?

Click to expand...

Yes!

b) Repeat part (a) with the requirement that
the first or second question, but not both, are answered correctly

Click to expand...

There are two cases:

[1] #1 is right, #2 is wrong.
. . .The other 8 questions can be answered in \(\displaystyle 2^8\) ways.

[2]#1 is wrong, #2 is right.
. . . The other 8 questions can be answered in \(\displaystyle 2^8\) ways.

Therefore, there are: \(\displaystyle \:2\,\times\,2^8\:=\:\fbox{512\text{ ways}}\)

c) Repeat part (a) in the case that three of the first five questions are answered correctly.

Click to expand...

For three right among the first five questions, there are \(\displaystyle \begin{pmatrix}5\\3\end{pmatrix}\) ways.
The other five questions can be answered in \(\displaystyle 2^5\) ways.

Therefore, there are: \(\displaystyle \,\begin{pmatrix}5\\3\end{pmatrix}\,\times\,2^5\:=\10)(32)\:=\:\fbox{320\text{ ways}}\)

Soroban, did you understand that in each of the three parts there are exactly eight questions answered correctly?