Consider all the permutations of the letters in the word MISSISSIPPI, where I is a vowel and the other letters are consonants.Quantity AThe number of permutations in which all the vowels are next to each otherQuantity BThe number of permutations in which all the consonants are next to each other

Sure, let's break it down:

Quantity A: The number of permutations in which all the vowels are next to each other

In the word MISSISSIPPI, there are 4 I's (vowels) and 7 consonants (M, S, S, S, S, P, P). If we want all the vowels to be together, we can treat them as a single entity. So, we have 8 entities to arrange (4 S's, 2 P's, 1 M, and 1 I).

The number of ways to arrange these 8 entities is 8!/(4!*2!) = 420 ways (we divide by 4! for the S's and 2! for the P's because they are identical and indistinguishable).

However, within the I entity, there are 4 I's that can be arranged in 4! = 24 ways.

So, the total number of permutations where all the vowels are together is 420 * 24 = 10,080.

Quantity B: The number of permutations in which all the consonants are next to each other

If we want all the consonants to be together, we can treat them as a single entity. So, we have 5 entities to arrange (4 I's, and 1 consonant entity).

The number of ways to arrange these 5 entities is 5! = 120 ways.

However, within the consonant entity, there are 7 consonants (4 S's, 2 P's, 1 M) that can be arranged in 7!/(4!*2!) = 1,260 ways.

So, the total number of permutations where all the consonants are together is 120 * 1,260 = 151,200.

Therefore, Quantity B is greater than Quantity A.