There is a word AUSTRALIA. Four letters are taken at rando

gmatter2012 wrote: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?

No. of vowels - 3As, 1U, 1I
No. of consonants - 1S, 1T, 1R, 1L

Two vowels and two consonants need to be chosen at a time.
1) CASE-1 If those two vowels are same i.e., 2As and two different consonants are chosen in 4C2 (6 ways).
No. of permutations for those two chosen consonants and vowels for a 4 letter word - 4!/2! = 12 ways
As we can choose the consonants in 6 ways, total no. of different words = 72
2) CASE-2: Both vowels are different and both consonants are different.
We can choose 2 different vowels in 3 ways (A,U), (U,I), (A,I)
We can choose 2 different consonants in 4C2 ways = 6 ways
To form a 4 letter word, (for any 2 selected vowels and any 2 selected consonants) No. of permutations will be = 4! = 24 ways.
Total no. of 4 letter words = 6*3*24 = 432
Adding the two case, Total no. of 4 letter words = 432+72= 504

Kindly write the OA. I am not very sure about my approach here

Why cant we consider (A,A) as a set?
I am little confused ....

avinashrao9 wrote:Why cant we consider (A,A) as a set?
I am little confused ....

We did take that set into consideration in case 1). Please take a look at the solution I provided.
Only after OA, I can expound this problem further.

kanwar86 wrote:

avinashrao9 wrote:Why cant we consider (A,A) as a set?
I am little confused ....

We did take that set into consideration in case 1). Please take a look at the solution I provided.
Only after OA, I can expound this problem further.

I was too ignorant while posting it and was later searching on how to delete it
A big blunder. Hope I dont do such things during the exam

kanwar86 wrote:

gmatter2012 wrote: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?

No. of vowels - 3As, 1U, 1I
No. of consonants - 1S, 1T, 1R, 1L

Two vowels and two consonants need to be chosen at a time.
1) CASE-1 If those two vowels are same i.e., 2As and two different consonants are chosen in 4C2 (6 ways).
No. of permutations for those two chosen consonants and vowels for a 4 letter word - 4!/2! = 12 ways
As we can choose the consonants in 6 ways, total no. of different words = 72
2) CASE-2: Both vowels are different and both consonants are different.
We can choose 2 different vowels in 3 ways (A,U), (U,I), (A,I)
We can choose 2 different consonants in 4C2 ways = 6 ways
To form a 4 letter word, (for any 2 selected vowels and any 2 selected consonants) No. of permutations will be = 4! = 24 ways.
Total no. of 4 letter words = 6*3*24 = 432
Adding the two case, Total no. of 4 letter words = 432+72= 504

Kindly write the OA. I am not very sure about my approach here

Bravo Kanwar that was amazingly solved !! appreciate it buddy!!
Your Logic is perfect .
Unfortunately I don't have the OA , but I am sure this must be the OA.