Helpppppp w/ Combination Problem?

Question

Tell how many ways two of Beethoven's nine symphonies can be chosen for a concert program for each situation. (Order's not important)

a. The Ninth symphony can't be chosen.

Would be 8C2? (C stands for Combination)

b. The Second Symphony must be chosen.

Well, since it's a group of two, i decided to look at it in a logical way. If the Second Symphony is chosen, then there's only one more Symphony to pick to create the group of two. Therefore, the answer would be 8. However, I was wondering how I would do it if there was a larger group than 2 (a group which would be difficult to answer it logically).

c. The Ninth Symphony can't be paired with the third, sixth, or seventh.

First, I found the total # of the groups of two, and got 9C2. I'm not sure what to do next...

thank you in advance ^^

Answer 1

Prob and combinitorics is not my stong suit, but I might be able to help a little.
Your first answer is correct. Basically you have 8 things and you want the combos two at a time since you're not interested in order.
Your second one is correct too. To answer your question about a group more than 2; Say you wanted 3 symphonies and one of them has to be the second. That leaves 8 and you want the combos of them taken 2 at a time.
Again you're right on the third part. Take 9C2 and subract 3 since the three combos of the third, sixth and seventh combined with the ninth are not allowed.
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Answer 2

a. The Ninth symphony can't be chosen.

Therefore you can only choose 2 from the remaining 8.
8C2 = 56/2 = 28
You are correct

b. The Second Symphony must be chosen.
1*(9-1) = 1*8 = 8
You are correct

c. The Ninth Symphony can't be paired with the third, sixth, or seventh.

Break it into two groups--one in which the ninth is chosen and one in which it isn't. Then add.

8C2 + 1(9 - 4) = 8C2 + 5 = 28 + 5 = 33

Or your way. Subtract the disallowed combinations.

9C2 - 1*3 = 36 - 3 = 33