Question

How many words can we build using exactly 5 A's, 5 B's and 5 C's if the first 5 letters cannot be A's, the second 5 letters cannot be B's and the third 5 letters cannot be C's? Hint: Group the different ways according to the number of B's in the first group.

Answer 1

Answer on Question #56773 – Math – Combinatorics | Number Theory

How many words can we build using exactly 5 A's, 5 B's and 5 C's if the first 5 letters cannot be A's, the second 5 letters cannot be B's and the third 5 letters cannot be C's? Hint: Group the different ways according to the number of B's in the first group.

Solution

Assume that exactly $N$ A's stand on the last 5 positions (so $N \in \{0,1,2,3,4,5\}$). It follows that there are $5 - N$ B's on the last 5 positions and $5 - N$ A's on the middle 5 positions. Also it follows that there are $N$ C's on the middle 5 positions and $5 - N$ C's on the first 5 positions.

So there are $\binom{5}{N}$ ways to place $N$ A's on the last 5 positions. Then there are $\binom{5}{5-N}$ ways to place the rest A's on the middle 5 positions. Finally, there are $\binom{5}{N}$ ways to place $N$ B's on the first 5 positions. So if we place $N$ A's on the last 5 positions, $5 - N$ A's on the middle 5 positions and $N$ B's on the first 5 positions, then the rest positions can be filled in the only way – C's on the first and on the middle 5 positions, B's on the last 5 positions.

Hence the number of words with $N$ A's on the last 5 positions is equal to

$$\binom{5}{N} \cdot \binom{5}{5 - N} \cdot \binom{5}{N} = \binom{5}{N}^{3}$$

and the total number is

$$\sum_{N=0}^{5} \binom{5}{N}^{3} = 1^{3} + 5^{3} + 10^{3} + 10^{3} + 5^{3} + 1^{3} = 2252.$$

Answer: 2252.

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