+0  
 
0
56
2
avatar+79 

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?

 Apr 25, 2020
 #1
avatar
+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.

 Apr 25, 2020
 #2
avatar+109527 
+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 :)

 Apr 25, 2020

13 Online Users

avatar