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Consider the word MULTIPLE and let a, b, c respectively be the number of ways of arranging letters of the word MULTIPLE such that
i) order of the vowels is not changed
ii) Positions of each of the vowels is fixed
iii) relative order of vowels and consonants is not changed. Then

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a = 3360

b = 60

b = 84

c = 216

answer is A, C.

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Detailed Solution

$a =^{8} C_{3} \times \frac{5!}{2!} = 3360$ (select 3 places for U,I,E and arrange remaining)

$b = \frac{5!}{2!}$ (U in 2^nd place, I in 5^th place and E in last place)

$c = 3! \times \frac{5!}{2!} = 360$

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