Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Question

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 0, which are divisible by 5 and none of the digits is repeated?

Answer 1

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
= 210.

So number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters and number of ways of arranging 5 letters among themselves
= 5!
= 120.

Required number of ways = (210 x 120) = 25200.

Answer 2

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

first we select 3 consonants from 7 consonants:
By different ways
7C3=35,
and
2 vowels from 4 vowels
4C2=6,
so total ways of selecting 2 vowels from 4 vowels and 3 consonants from 7 consonants =35*6=210.
and we can arrange 5 letters 5! ways, so total ways =210*5!
=210*120
=25200.

Answer 3

Answer is 1,764,000

The 4 different consonants can be selected in 7C4 ways
The different vowels can be selected in 5C3 ways
The resulting 7 different letters ( 4 consonants & 3 vowels) can be arranged themselves in 7P7 = 7! ways..

So the answer is 7C4 * 5C3 * 7! = 35 *10 * 5040 = 1764000

Comments

Before solving any question read the Question atleast twice.

It is asked that to choose 3 consonants out of 7 not 4.