What is the probability that the two E's come | vedclass

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Question

(A) The word $PEACE$ consists of $5$ letters: $P, E, A, C, E$. The total number of arrangements of these letters is $\frac{5!}{2!} = \frac{120}{2} = 6

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$2/5$

Vedclass · Answer

(A) The word $PEACE$ consists of $5$ letters: $P, E, A, C, E$. The total number of arrangements of these letters is $\frac{5!}{2!} = \frac{120}{2} = 60$. To find the number of arrangements where both $E$'s come together,we treat the two $E$'s as a single unit $(EE)$. Now we have $4$ units: $P, A, C, (EE)$. The number of ways to arrange these $4$ units is $4! = 24$. The required probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{24}{60} = \frac{2}{5}$.

What is the probability that the two $E$'s come together in the words formed by the letters of the word $PEACE$?

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