Can you help me with this simple probability question
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Can you help me with this simple probability question

[From: ] [author: ] [Date: 12-11-20] [Hit: ]
the two Is can occupy 11C2 = 55 different positions.Out of these 55, 10 of them have the two Is next to each other.Therefore, 55 - 10 = 45 positions where the two Is are NOT next to each other.Next,......
What is the probability of arranging the letters of the word PROBABILITY at random so that the two I's are separate?
Please explain.
Thanks

-
Joshua

There are 11 letters. In general, the two I's can occupy 11C2 = 55 different positions. Out of these 55, 10 of them have the two I's next to each other. Therefore, 55 - 10 = 45 positions where the two I's are NOT next to each other.

Next, we need to fill the remaining 9 positions with 9 letters of which 2 are the same (B,B). There are 9! / (2!7!) = 36 ways to assign the remaining 9 letters.

So, there are 45 x 36 = 1620 ways of arranging the letters of the word PROBABILITY at random so that the two I's are separate

In general, there are 11! / (2!2!7!) = 1980 ways to assign 11 letters of which there are 2 pairs the same (I,I,B,B)

Finally, the probability = 1620 / 1980 = 9/11

Hope that helped
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