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Find the number of seven-letter words that use letters from the set $\{\alpha, \beta, \gamma\}$ and contain at least one $\alpha$ and at least two $\beta$'s.

I posted a similar problem before, in which one had to find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$ and contain at least one each of $\alpha$, $\beta$ and $\gamma$. How can I modify that answer to the question above?

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  • $\begingroup$ Using $a,b,c$ would need less typing. $\endgroup$ – John Bentin Aug 4 '17 at 9:55
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There are $\binom{7}{k}$ ways to fill exactly $k$ of the seven positions with a $\beta$ and $2^{7 - k}$ ways of filling the remaining positions with one of the other two letters. Of these $2^{7 - k}$ ways of filling the remaining positions with $\alpha$ or $\gamma$, only one way involves using only $\gamma$. Hence, there are $$\binom{7}{k}(2^{7 - k} - 1)$$ ways to fill exactly $k$ of the positions with $\beta$s if at least one $\alpha$ is used. Thus, the number of seven-letter words with at least two $\beta$s and at least one $\alpha$ is $$\sum_{k = 2}^{7} \binom{7}{k}(2^{7 - k} - 1) = \binom{7}{2}(2^5 - 1) + \binom{7}{3}(2^4 - 1) + \binom{7}{4}(2^3 - 1) + \binom{7}{5}(2^2 - 1) + \binom{7}{6}(2^6 - 1) + \binom{7}{7}(2^0 - 1) = 1491$$

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  • Strategy similar to your first question:

Number of 7-letters words: $|U|=3^7$

Number of 7-letters words without $\alpha$ (resp.$\beta$): $|A|=|B|=2^7$

Number of 7-letters words without $\alpha$ nor $\beta$: $|AB|=1$

Number of 7-letters words with one and only one $\beta$ and no $\alpha$ : $|AB1|=7$

Number of 7-letters words with one and only one $\beta$ : $|B1|=7*2^6$

Answer to the question: $|U|-|A|-|B|+|AB|-|B1|+|AB1|=3^7-2*2^7+1-7*2^6+7=1491$

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  • $\begingroup$ You have to choose positions for the one $\beta$. $\endgroup$ – N. F. Taussig Aug 4 '17 at 9:48
  • $\begingroup$ Oh yes, thx @N.F.Taussig $\endgroup$ – Evargalo Aug 4 '17 at 9:53
  • $\begingroup$ I'm a little confused. How did you get $|AB1|=7$ and $|B1|=7*2^6$ @Evargalo? Are we assuming that $\alpha$ and two $\beta$s are missing for $|AB1|$? $\endgroup$ – Nixie777 Aug 5 '17 at 10:12
  • $\begingroup$ @Nixie777 If a seven-letter word has exactly one $\beta$ and no $\alpha$, it consists of one $\beta$ and six $\gamma$s. Choosing which of the seven positions is filled with $\beta$ completely determines the word. Hence, $|AB1| = 7$. If a seven-letter word has exactly one $\beta$, then we can choose its position in seven ways. The remaining six positions can be filled with either $\alpha$ or $\gamma$. Hence, $|B1| = 7 \cdot 2^6$. $\endgroup$ – N. F. Taussig Aug 5 '17 at 10:54
  • $\begingroup$ @N.F.Taussig thank you. That makes more sense now. $\endgroup$ – Nixie777 Aug 5 '17 at 11:01

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