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Question:

How many words which can be formed with INSTITUTION such that vowels and consonants are alternate?

My Attempt:

There are total 11 letters in word INSTITUTION. The 6 consonants are {NSTTTN} and the 5 vowels are {IIUIO}. So if we begin with consonant then we can have $6!$ different arrangement of consonants and $5!$ different arrangement for vowels. But I is repeated 3 times, T is repeated 3 times and N is repeated 2 times. Thus we get $$\frac{6! \cdot 5!}{3! \cdot 3! \cdot 2!} = 1200$$ different words.

Now, It is also possible that that word begins with a vowel, thus we will have another $1200$ words.

Thus total number of words formed is $1200 + 1200 = 2400$

But answer given is 1200.

Am I missing something?

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You do not have the option of either starting with a vowel or starting with a consonant since you must alternate and you have 6 consonants and 5 vowels. It must start with a consonant.

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  • $\begingroup$ Oh okay I see INSTITUTION have only 5 vowels! well thanks for pointing it out :-) $\endgroup$ – Freddy May 22 '15 at 17:30
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It is NOT possible that the word begins with a vowel, because there are more consonants.


The number of ways to place NNSTTT in the odd places is:

$$\binom62\cdot\binom41\cdot\binom33=60$$

The number of ways to place IIIOU in the even places is:

$$\binom53\cdot\binom21\cdot\binom11=20$$

Hence the number of ways to arrange INSTITUTION is:

$$60\cdot20=1200$$

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If a word starts with a vowel, we will run out of vowel before we run out of consonant.

Let C denote a consonant, and V a vowel, we have in the first case CVCVCVCVCVC, but in the second case we will have VCVCVCVCVC and we still have one C left. Which we can not add at the end.

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