If the letters of the word INTERMEDIATE are a | vedclass

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(A) The word $INTERMEDIATE$ has $12$ letters: $I(2), N(1), T(2), E(3), R(1), M(1), D(1), A(1)$.Total arrangements $= \frac{12!}{2! 	imes 2! 	imes 3!

Vedclass · Accepted Answer

$6/11$

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(A) The word $INTERMEDIATE$ has $12$ letters: $I(2), N(1), T(2), E(3), R(1), M(1), D(1), A(1)$.Total arrangements $= \frac{12!}{2! 	imes 2! 	imes 3!}$.To find the probability that the three $E$'s do not come together,we first arrange the other $9$ letters: $I, I, N, T, T, R, M, D, A$. The number of ways to arrange these is $\frac{9!}{2! 	imes 2!}$.There are $10$ gaps created by these $9$ letters. We can place the $3$ $E$'s in these $10$ gaps in $^{10}C_3$ ways.Number of favorable arrangements $= \frac{9!}{2! 	imes 2!} 	imes ^{10}C_3$.Probability $= \frac{\frac{9!}{2! 	imes 2!} 	imes ^{10}C_3}{\frac{12!}{2! 	imes 2! 	imes 3!}} = \frac{9! 	imes 120 	imes 3!}{12!} = \frac{120 	imes 6}{12 	imes 11 	imes 10} = \frac{720}{1320} = \frac{6}{11}$.

If the letters of the word $INTERMEDIATE$ are arranged,what is the probability that the two $E$'s do not come together?

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