Letters in the word HULULULU are rearranged. | vedclass

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(C) The word $HULULULU$ contains $8$ letters: $H(1), U(4), L(3)$.Total number of arrangements $n(S) = \frac{8!}{4!3!} = \frac{40320}{24 	imes 6} = \f

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$\frac{3}{28}$

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(C) The word $HULULULU$ contains $8$ letters: $H(1), U(4), L(3)$.Total number of arrangements $n(S) = \frac{8!}{4!3!} = \frac{40320}{24 	imes 6} = \frac{40320}{144} = 280$.To find the number of arrangements where all three $L$ are together,we treat $(LLL)$ as a single unit.Now we have $6$ units: $(LLL), H, U, U, U, U$.The number of arrangements $n(A) = \frac{6!}{4!1!} = \frac{720}{24} = 30$.The required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{30}{280} = \frac{3}{28}$.

Letters in the word $HULULULU$ are rearranged. The probability of all three $L$ being together is

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