Anonymous wrote:

There is a word AUSTRALIA. Four letters are taken at random. What is the probability to have two vowels and two consonants?

Vowels: A, U, A, I, A.

Consonants: S, T, R, L

This is a case of without replacement.

Without replacement:

Total outcomes: C(9,4)=9*8*7*6/4!=126

Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60

Probability=60/126=10/21

But we can also consider the case of with replacement:

Total outcomes: 9^4

Outcomes with two vowels and two consonants: P(4,2)*5^2*4^2

You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above question the repeating letters don't matter. However they would matter is the question is asked differently.

Example:

There is a word AUSTRALIA. Four letters are taken at random. How many

different words can be formed that have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.

Consonants: S, T, R, L

Outcomes with two vowels and two consonants:

1) The two vowels are different: C(3,2)*C(4,2)

2) The two vowels are the same: C(1,1)*C(4,2)

Total outcomes = 18+6=24