Question

How many words can be formed by taking $3$ consonants and $2$ vowels out of $5$ consonants and $4$ vowels.
A) ${}^5{{\text{C}}_3} \times {}^4{{\text{C}}_2}$
B) $\dfrac{{{}^5{{\text{C}}_3} \times {}^4{{\text{C}}_2}}}{5}$
C) ${}^5{{\text{C}}_3} \times {}^4{{\text{C}}_3}$
D) $\left( {{}^5{{\text{C}}_3} \times {}^4{{\text{C}}_2}} \right)\left( {5} \right)!$

Answer 1

We use the combinations concept to choose $5$ letters ($2$ vowels, $3$consonants) out of $9$ letters ($5$consonants, $4$ vowels). After choosing the letter we use Permutation concept to ARRANGE selected $5$ letters.

Complete Step-by-step Solution
$5$ consonants and $4$ vowels are given out of which a $5$ letter word is to be formed and only $3$ consonants and $2$ vowels can be used.
We have to select $3$ consonants out of $5$ consonants and $2$ vowels out of $4$ vowels.
We can select 3 consonants out of $5$ in ${}^5{{\text{C}}_3}$ways. (Apply the concept of combination here).
We can select $2$ vowels out of $4$ vowels in ${}^4{{\text{C}}_2}$ ways. (Apply the concept of combination here).
And also a $5$ letter word can be formed in $5!$ ways. (Apply the concept of permutation here)
Now, we will have to multiply all the ways we found to get the required ways
So total words can be formed $= \left( {{}^5{{\text{C}}_3} \times {}^4{C_2}} \right)\left( {5} \right)!$

So option (D) which is $\left( {{}^5{{\text{C}}_3} \times {}^4{C_2}} \right)\left( {5} \right)!$ is the correct answer.

Note:
As we will have to form a word of $5$ letter and the letter must not be repeated that is why we will multiply the selected ways by $5!$ and $5! = 5 \times 4 \times 3 \times 2 \times 1$ ways because we will have to form a word of $5$ letters and none of the letters will be repeated.