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The number of ways of arranging the letters of the word ‘SUCCESSFUL’ so that no two S’s are together is

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$\frac{7!}{2! 2!}$

$\frac{7!}{2! 2!} \frac{^{8} P_{3}}{3!}$

$\frac{7!}{2! 2!}$

$\frac{^{8} P_{3}}{3!}$

answer is B.

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Detailed Solution

Remaining 7 letters (excluding of 3'S') arranged as $\frac{7!}{2! 2!}$ ways and these arrangement having 8 empty places then

these 8 empty places arranged 3'S' therefore

Total arrangement= $\frac{7!}{2! 2!} \cdot 8 c_{3}$

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