Question
The number of permutations that can be made out of the letters of the word “EQUATION” which start with a consonant and end with a consonant is
Answer: Option B
:
B
Consonants occupy 2 ends in 3P2ways remaining 6 letters occupy 6 places in 6! Ways
So the required number of arrangements = 3P2.6!=3!6!
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:
B
Consonants occupy 2 ends in 3P2ways remaining 6 letters occupy 6 places in 6! Ways
So the required number of arrangements = 3P2.6!=3!6!
Was this answer helpful ?
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