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(A) The word $'UNIVERSITY'$ contains $10$ letters, where the letter $I$ appears $2$ times.Total number of permutations $= \frac{10!}{2!}$.To find the

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$\frac{4}{5}$

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(A) The word $'UNIVERSITY'$ contains $10$ letters, where the letter $I$ appears $2$ times.Total number of permutations $= \frac{10!}{2!}$.To find the number of arrangements where two $I$s are together, we treat the two $I$s as a single unit. Now, we have $9$ units (the block $(II)$ and the other $8$ letters), which can be arranged in $9!$ ways.Number of arrangements where two $I$s are together $= 9!$.Number of arrangements where two $I$s are not together $= 	ext{Total arrangements} - 	ext{Arrangements where } I	ext{s are together}$.$= \frac{10!}{2} - 9! = \frac{10 	imes 9!}{2} - 9! = 9!(5 - 1) = 4 	imes 9!$.Required probability $= \frac{4 	imes 9!}{\frac{10!}{2}} = \frac{4 	imes 9! 	imes 2}{10!} = \frac{8 	imes 9!}{10 	imes 9!} = \frac{8}{10} = \frac{4}{5}$.

Find the probability that in a random arrangement of letters of the word $'UNIVERSITY'$, two $I$s do not come together.

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