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Question: How many ways are there to arrange the letters of the word COMPUTE so that vowels OR consonants are in alphabetical order OR at least two vowels next to each other.

I have tried lots of things to solve this question. I have started by looking for the number of permutations where vowels are in alphabetical order. (7!/4!) Then, I looked for the number of permutations where all the consonants in alphabetical order. (7!/3!) Then I subtracted the number of arrangements where all the consonants and vowels are in alphabetical order. (C(7,3)). That is the point I stuck. How should I include the condition of having "OR at least two vowels next to each other." ?

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  • $\begingroup$ so that vowels or consonants are in alphabetical order? $\endgroup$ – VortexYT Dec 3 '20 at 21:01
  • $\begingroup$ yes, I've add "are". Thank you. $\endgroup$ – yunusemre Dec 3 '20 at 21:22
  • $\begingroup$ You seem to have confused permutations with vowels in alphabetical order with those with consonants in alphabetical order. For vowels in alphabetical order, there are seven ways to place the C, six ways to place the M, five ways to place the P, and four ways to place the T. The three vowels can only be placed in the remaining three positions in one way. Hence, there are $7 \cdot 6 \cdot 5 \cdot 4 = \frac{7!}{(7 - 4)!} = \frac{7!}{3!}$ permutations in which the vowels appear in alphabetical order. $\endgroup$ – N. F. Taussig Dec 3 '20 at 22:52
  • $\begingroup$ A similar argument shows that there are $\frac{7!}{4!}$ permutations in which the vowels appear in alphabetical order. $\endgroup$ – N. F. Taussig Dec 3 '20 at 22:52
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Start by counting the number of ways to choose the positions of the vowels: $\binom{7}{3}=35$

Then find how many of these do not have vowels in consecutive positions. With four consonants, there are five places to insert vowels: _c_c_c_c_ Choose three of them: $\binom{5}{3}=10$

So there are $35-10=25$ patterns that meet the condition of having consecutive vowels. For each of these there are $4!$ permutations of the consonants and $3!$ permutations of the vowels. This gives a total of $25\times 4!\times 3!=3600$ permutations with at least two vowels next to each other.

For the $10$ patterns that do not have consecutive vowels, there are $4!$ permutations with the vowels in order and $3!$ with the consonants in order. The permutation with both groups in order is counted twice, so there are $4!+3!-1=29$ permutations with at least one group in alphabetical order. This gives a total of $10\times 29=290$ additional permutations that meet the given conditions.

The total is therefore $3890$ arrangements.

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  • $\begingroup$ Thank you for your answer. However, isn't the number of total permutations 7!=5040? Maybe I should have added "without repetition" condition. $\endgroup$ – yunusemre Dec 8 '20 at 10:19
  • $\begingroup$ @yunusemre Yeah, my original result was insane. I have corrected the error. $\endgroup$ – Daniel Mathias Dec 8 '20 at 22:55

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