To solve this problem, we first consider the two I's as a single entity. So, the word "UNIVERSITY" becomes "UN_VERS_TY" (where _ represents the two I's together).

This gives us 9 entities to arrange (7 letters + 1 pair of I's), which can be done in 9! ways.

However, we have to account for the repetition of the letters 'U' and 'N'. So, we divide by 2! for each of these repetitions to avoid over-counting.

This gives us a total of 9! / (2! * 2!) ways to arrange the letters such that the I's are together.

The total number of arrangements of the word "UNIVERSITY" without any restrictions is 10! / (2! * 2! * 2!) (10 letters, with 'I', 'U', and 'N' each repeating twice).

So, the probability is (9! / (2! * 2!)) / (10! / (2! * 2! * 2!)) = 1/10.

None of the options A, B, C, D match this result. There might be a mistake in the question or the options provided.

Question

To solve this problem, we first consider the two I's as a single entity. So, the word "UNIVERSITY" becomes "UN_VERS_TY" (where _ represents the two I's together).

This gives us 9 entities to arrange (7 letters + 1 pair of I's), which can be done in 9! ways.

However, we have to account for the repetition of the letters 'U' and 'N'. So, we divide by 2! for each of these repetitions to avoid over-counting.

This gives us a total of 9! / (2! * 2!) ways to arrange the letters such that the I's are together.

The total number of arrangements of the word "UNIVERSITY" without any restrictions is 10! / (2! * 2! * 2!) (10 letters, with 'I', 'U', and 'N' each repeating twice).

So, the probability is (9! / (2! * 2!)) / (10! / (2! * 2! * 2!)) = 1/10.

None of the options A, B, C, D match this result. There might be a mistake in the question or the options provided.

Knowee AI · Accepted Answer

To solve this problem, we first consider the two I's as a single entity. So, the word "UNIVERSITY" becomes "UN_VERS_TY" (where _ represents the two I's together).

This gives us 9 entities to arrange (7 letters + 1 pair of I's), which can be done in 9! ways.

However, we have to account for the repetition of the letters 'U' and 'N'. So, we divide by 2! for each of these repetitions to avoid over-counting.

This gives us a total of 9! / (2! * 2!) ways to arrange the letters such that the I's are together.

The total number of arrangements of the word "UNIVERSITY" without any restrictions is 10! / (2! * 2! * 2!) (10 letters, with 'I', 'U', and 'N' each repeating twice).

So, the probability is (9! / (2! * 2!)) / (10! / (2! * 2! * 2!)) = 1/10.

None of the options A, B, C, D match this result. There might be a mistake in the question or the options provided.

What is the probability of the random arrangement of letters in the word "UNIVERSITY" such that the two I's should be together? A. 1/5 B. 3/5 C. 2/7 D. 1/7

Question

What is the probability of the random arrangement of letters in the word "UNIVERSITY" such that the two I's should be together?

Solution

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