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(B) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.Total arrangements $= \frac{12!}{3! 	imes 2! 	imes 4!} = \frac{47

Vedclass · Accepted Answer

$1646400$

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(B) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.Total arrangements $= \frac{12!}{3! 	imes 2! 	imes 4!} = \frac{479001600}{6 	imes 2 	imes 24} = 1663200$.Treating all vowels $(I, E, E, E, E)$ as one unit,we have $8$ consonants $(N, N, N, D, D, P, C)$ and $1$ vowel unit. Total units $= 9$.Arrangements with vowels together $= \frac{9!}{3! 	imes 2!} 	imes \frac{5!}{4!} = \frac{362880}{6 	imes 2} 	imes 5 = 30240 	imes 5 = 151200$.Number of arrangements where vowels never occur together $= 1663200 - 151200 = 1512000$.

Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do the vowels never occur together?

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