Find the probability that in a random arrange | vedclass

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Question

(A) The word $DAUGHTER$ contains $8$ distinct letters $(D, A, U, G, H, T, E, R)$.The total number of ways to arrange these $8$ letters is $n(S) = 8!$.

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(A) The word $DAUGHTER$ contains $8$ distinct letters $(D, A, U, G, H, T, E, R)$.The total number of ways to arrange these $8$ letters is $n(S) = 8!$.If the letter $D$ is fixed at the first position,we need to arrange the remaining $7$ letters in the remaining $7$ positions.The number of favorable outcomes is $n(A) = 7!$.Therefore,the required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{7!}{8!} = \frac{7!}{8 	imes 7!} = \frac{1}{8}$.

Find the probability that in a random arrangement of the letters of the word $DAUGHTER$,the letter $D$ occupies the first place.

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