The letters of the word ASSASSIN are written | vedclass

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(B) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(1), N(1)$.Total number of arrangements $ = \frac{8!}{4! 	imes 2!} = \frac{8 	imes 7 	i

Vedclass · Accepted Answer

$\frac{1}{14}$

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(B) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(1), N(1)$.Total number of arrangements $ = \frac{8!}{4! 	imes 2!} = \frac{8 	imes 7 	imes 6 	imes 5 	imes 4!}{4! 	imes 2} = 840$.To ensure no two $S$ occur together,we use the gap method. First,arrange the remaining $4$ letters $(A, A, I, N)$.The number of ways to arrange these $4$ letters is $\frac{4!}{2!} = 12$.These $4$ letters create $5$ gaps (including ends): $\_ L_1 \_ L_2 \_ L_3 \_ L_4 \_$.We need to place $4$ $S$'s in these $5$ gaps. The number of ways to choose $4$ gaps out of $5$ is $\binom{5}{4} = 5$.Total favorable arrangements $ = 12 	imes 5 = 60$.Probability $ = \frac{	ext{Favorable arrangements}}{	ext{Total arrangements}} = \frac{60}{840} = \frac{1}{14}$.

The letters of the word $ASSASSIN$ are written down at random in a row. The probability that no two $S$ occur together is

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