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?
Best Answer
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.