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
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
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