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Part a): Find the number of eight-letter words that use letters from the set {A, B, C} and contain exactly three As. (Hint: Begin by selecting positions for the As.)

My attempt: There are $\binom{8}{3}$ ways to place the As. As for the rest of the letters?

The attempt above is incorrect. This was just my scratch work.

Part b): How many of the words counted in part a) contain no two consecutive As?

Basically, I need help with both part, and I think using the combination numbers above will help.

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Find the number of eight-letter words that use letters from the set $\{A, B, C\}$ and contain exactly three A's.

There are $\binom{8}{3}$ ways to choose the positions for the A's. Each of the five remaining positions can be filled with a B or a C. Hence, the number of such words is $$\binom{8}{3}\cdot 2^5$$

How many of these words contain no two consecutive A's?

Consider a five-letter word composed of B's and C's such as $$BBCBC$$ We wish to insert three A's so that no two of them are consecutive. We have six spaces in which to insert the three A's, indicated by the squares in the example. $$\square B \square B \square C \square B \square C \square$$ To ensure that no two A's are consecutive, we must choose three of these six spaces into which to insert an A. For instance, if we choose the first, second, and fifth spaces, we obtain the sequence $$ABABCBAC$$

There are $2^5$ five-letter words composed of B's and C's and $\binom{6}{3}$ ways to choose three of the six spaces in which to insert an A. Hence, there are $$2^5 \cdot \binom{6}{3}$$ eight-letter words composed from the letters A, B, and C with exactly three A's in which no two of the A's are consecutive.

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For every one of the $\binom 83$ ways of placing the A's there are $2^5$ ways of filling the remaning five places with B or C.

Out of the $\binom 8 3$ ways of placing A's, there are 6 ways of having all three together and $2\binom 62$ ways of placing a single A and a group of 2 A's.

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  • $\begingroup$ $2\binom72$ ways to place a single $A$ and group of $2$ $A$s $\endgroup$ – Zach Teitler Jul 5 '17 at 4:27
  • $\begingroup$ @ZachTeitler I don't think so, there are 5 non A's so 6 places to put separated groups of A's $\endgroup$ – WW1 Jul 5 '17 at 4:42
  • $\begingroup$ Oh right. They have to be separated. Oops. $\endgroup$ – Zach Teitler Jul 5 '17 at 5:27

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