Four letters are taken at random from the word AUSTRALIA without repla

Question

Four letters are taken at random from the word AUSTRALIA. How many different words can be formed that have two vowels and two consonants?

A. 72
B. 216
C. 240
D. 432
E. 504

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From the word AUSTRALIA,

chetan2u · Accepted Answer

We have 9 alphabets, out of which there are 5 vowels and 4 consonants. There are 3 similar vowels while all consonants are distinct. 
Two case
1) All distinct alphabets
Choose 2 out of 3 distinct vowels and 2 out of 4 consonants => 3C2*4C2=3*6=18
But these 4 selected can be arranged in 4! ways = 18*24=432

2) 2 similar vowels and 2 distinct consonants 
2 out of 4 consonants => 4C2=6
But these 4 selected can be arranged in 4!/2! ways = 6*12=72

Total ways =432+72=504

E

madina02 · Answer

Hi, this is a great question.

I wonder if there's a way to solve it with the permutation formula?

Regor60 · Answer

3 distinct vowels, 4 consonants.

2 vowels can be selected

3!/2! = 3 ways

2 consonants can be selected

4!/2!2! = 6 ways

4 letters can be arranged

4! = 24 ways

24*3*6 = 432 ways.

This is the minimum number before considering selecting 2 A's.

Since 504 is the only option remaining, there is no need to calculate the effect of two A's.

Posted from my mobile device

HarshavardhanR · Answer

- Vowels: (a, a, a), u, i.
- Consonants: s, t, r, l

- The consonants are all unique, hence, when we pick two consonants, the two consonants will always be different.
- With the task of picking two vowels, there are two broad cases

a) When both vowels picked are a's. Here, we get repeated vowels.
b) When we pick two different vowels (say - a and u, u and i, etc.)

Let's count all the possibilities.

(1) Case 1: (v1, v2, c1, c2) - 2 different vowels, 2 different consonants

- #ways to choose 2 unique vowels from 3 distinct options (a, u, i) --- 3C2
- # ways to choose 2 unique consonants from 4 distinct options (s, t, r, l) --- 4C2
- #ways to arrange these 4 unique letters = 4!
-  Total #ways = 3C2 x 4C2 x 4! = 3 x 6 x 24 = 432.

(2) Case 2: (v, v, c1, c2) - 2 vowels being the same (i.e., 2 a's), 2 different consonants

- #ways to choose 2 a's from the 3 a's (a, a, a) --- 1 (identical letters, there are no "different" ways)
- # ways to choose 2 unique consonants from 4 distinct options (s, t, r, l) --- 4C2
- #ways to arrange these 4 letters = 4! / 2! (because the a's repeat)
-  Total #ways = 1 x 4C2 x 4!/2! = 1 x 6 x 12 = 72.
  
Final Answer = 432 + 72 = 504. Choice E.

Harsha

Four letters are taken at random from the word AUSTRALIA without repla

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Four letters are taken at random from the word AUSTRALIA without repla

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615 to 715 in 15 Days (Ria's Strategies)

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My Rewards