Question

How many 15 letter arrangements of 5 A's, 5 B's and 5 C's have no A's in the first 5 letters, no B's in the next 5 letters, and no C's in the last 5 letters.

Answer 1

Answer on Question #56761 – Math – Combinatorics | Number Theory

How many 15-letter arrangements of 5 A's, 5 B's and 5 C's have no A's in the first 5 letters, no B's in the next 5 letters, and no C's in the last 5 letters?

Solution

We have the following different cases, in each of them we use the formula $\overline{P_{n=k_1+k_2}}(k_1,k_2) = \frac{n!}{k_1! \cdot k_2!}$ :

1) The first 5 letters are 5 C's. Then the next 5 letters may be only all A's, and the last 5 letters may be only all B's. We have $(\overline{P_5}(5))^3 = 1$ arrangement.

2) In the first 5 letters there is only 1 letter B (other 4 letters may be only C's). Then in the second 5 letters there are 1 C and 4 A's, and in the last 5 letters there are 1 A and 4 B's. Total number of the different arrangements for this case is equal to $(\overline{P_5}(4,1))^3 = 5^3 = 125$.

3) In the first 5 letters there is only 2 letters B (other 3 letters may be only C's). Then in the second 5 letters there are 2 C's and 3 A's, and in the last 5 letters there are 2 A's and 3 B's. Total number of the different arrangements for this case is equal to $(\overline{P_5}(3,2))^3 = 10^3 = 1000$.

Other cases are the following:

4) In the first 5 letters there is only 3 letters B (other 2 letters may be only C's).

5) In the first 5 letters there is only 4 letters B (another letter may be only C).

6) In the first 5 letters there are 5 B's.

Cases 4)-6) will be symmetric to cases 3)-1).

Therefore, we have $(1 + 125 + 1000) \cdot 2 = 1126 \cdot 2 = 2252$ arrangements.

Answer: 2252.

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