permutation issue

Hi,

Here AAU is one set, C, B, S are others.
so total 4 sets can e arranged in 4! ways.
and the vowels internally can be arranged in 3!/2! ways(because A repeats twice)

so total ways is 4! * 3!/2! = 72

This one is easy enough to do by hand:

ABACUS has 3 vowels and 3 consonants. You want all the vowels together, so the possibilities are:

V V V C C C

C V V V C C

C C V V V C

C C C V V V

Now in the case of the vowels, there are only 3 unique arrangements, because there's a repeated letter:

AAU
AUA
UAA

(By the way, as a side note, there are usually n! unique arrangements of n elements, but when you have a repeated element, you have to divide by the factorial of the number of repeats. For example, this would be 3! unique arrangements, but because the A appears twice, it's actually 3! / 2! = 3. And if, for example, you had 7 elements, and 2 of them repeated twice, then it would bt 7! / (2! * 2!) )

In the case of the consonants, there are no repeats, so there are 3! = 6 unique arrangements. So:

V V V C C C (3 arrangements for the vowels) * (6 arrangements for the consonants) = 18 total arrangements

C V V V C C same thing, 3*6 = 18

C C V V V C Again, 18.

C C C V V V Again, 18.

Total arrangements: 72

Make sense?

ABACUS has 3 vowels AAU
now conside this as 1 unit so u have 4 alphabet which can be aranged 4! ways
secon case AAU can themselves arrange in 3!ways out of which there will be arrangement when 2 A will be together so AAU can arrange themself in 3!/2! =3 ways
so total number of ways 4!*3=72

make sense
4 elements

4!3! / 2! = 72