Permutation/Combinations in bit Strings I have a bit string with 10 letters, which can be {a, b, c}. How many bit strings can be made that have exactly 3 a's, or exactly 4 b's?
I thought that it would be C(7,2) + C(6,2), but that's wrong (the answer is 24,600).

Hint: By the inclusion-exclusion principle,  the answer is equal to 
$$\begin{align} & \text{(number of strings with exactly 3 a's)}\\ + & \text{(number of strings with exactly 4 b's)}\\ - &\text{(number of strings with exactly 3 a's and 4 b's)} \end{align}$$
Suppose I want to make a string with exactly 3 a's. First, I need to choose where to put the a's; the number of ways of choosing 3 places to put the a's, out of 10 places, is $\binom{10}{3}$. Now, I need to choose how to fill in the other places with b's or c's; there are 2 choices of letters and 7 places left. Thus, the number of strings that have exactly 3 a's is equal to
$$\binom{10}{3}\cdot 2^7$$
You should be able to use similar reasoning to find the other numbers.
