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  4. If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is 9! (k), then k equals

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Q. If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is $9! (k)$, then k equals

Permutations and Combinations

Solution:

$O = 3, M = 1, C = 1, Y = 1, N = 2, T = 2, I = 2$
$\frac{9 !}{2 ! \cdot 2 ! \cdot 2 !} \times{ }^{10} C _3=15 \times 9 ! $
$\Rightarrow k =15$