The word HIGHMOON has different permutations. These permutations can be represented by variable S. What is the probability that the two “H” won’t be together in a permutation?

Coolio Apr 25, 2020

#1**+1 **

Here is my attempt at this:

The word "HIGHMOON" has 8 letters. If you allow all letters to be permuted, in which case you will have "duplicates", you would get:

8! =40,320 permutations.

If you exclude duplicates, then you would have:

8! / 2!.2! =10,080 pemutations

However, in either case, you will have 2 "Hs" in your permutations, and since each letter appears an equal number of times in the total of ALL permutations, it therefore follows that in 2/8 =1/4 of ALL permutations, you will have 2 "Hs" next to each other. The other 6/8 = 3/4 of ALL permutations, they will NOT be next to each other. So, the answer is, I think, that in 3/4 or 75% of ALL permutations, they will NOT be together.

Note: somebody should check this. Thanks.

Guest Apr 25, 2020

#2**+1 **

HIGHMOON

IGMN OO HH

Number of permutations = 8!/2!2! = 10080

Number where the 2 hs are together is 7!/2! = 2520

resultant prob where they are together

\(=\frac{7!}{2!} \div \frac{8!}{2!2!}\\ =\frac{7!}{2!} \times \frac{2!2!}{8!}\\ =\frac{1}{1} \times \frac{2!}{8}\\ =\frac{1}{4} \)

So the prob that they are NOT together is 3/4

Just like guest said :)

Melody Apr 25, 2020