The letters ABCDEFGH are to be used to form strings of length 4. How many strings contain the letters A and B if repetitions are not allowed?
Here is my answer:
The total number of possible combinations are $8×7×6×5$
The total number of combinations if A and B are not allowed is $6×5×4×3$
So the answer is the difference of the two, which is $(8×7×6×5) – (6×5×4×3) = 1320$
Please tell me if I am correct and if not, point me in the right direction.
Best Answer
If the four-letter string must contain A and B, then there are $\left(6 \atop 2 \right) = 15$ pairs of letters you could choose from the remaining six letters C, D, E, F, G and H.
You now have $15$ possible sets of four letters. Each set of four letters can be arranged in $4! = 24$ different ways, giving you $15 \cdot 24 = 360$ possible strings.