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If all the L‘s occur together and also all I‘s occur together, when the letters of the word ‘HALLUCINATION’ are permuted, then the number of such arrangements of letters is: 

(a) \(\frac{7!}{2!2!2!2!}\)

(b) \(\frac{11!}{2!2!}\)

(c) \(\frac{13!}{2!2!2!2!}\)

(d) \(\frac{11!}{2!2!2!2!}\)

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(b) \(\frac{11!}{2!2!}\)

In the word ‘HALLUCINATION’ there are total 13 letters out of which 7 are consonants and 6 are vowels. Also there are 2L’s, 2N’s, 2A’s and 2I’s. 

If L’s occur together and I’s occur together, then we consider them as one unit each. So, the arrangement can be written as: 

Now these 11 letters (9 letters and 2 units of L and I) can be arranged in \(\frac{11!}{2!2!}\) ways. ( There are 2A’s and 2N’s)

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