Question

Total no. of words formed by using 2 vowels and 3 consonants taken from 4 vowels and 5 constants in equal to:
A. 60
B. 120
C. 720
D. None of these

Answer 1

Hint: This problem deals with permutations and combinations. But here a simple concept is used. Although this problem deals with combinations only. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$ \Rightarrow n! = n(n - 1)(n - 2).......1$
The no. of combinations of $n$ objects taken $r$ at a time is determined by the formula which is used:
$ \Rightarrow {}^n{c_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step-by-step answer:
Given that there are 4 vowels and 5 consonants.
We are asked to find the total no. of words formed by the words taken 2 vowels from the given 4 vowels and 3 consonants from the given 5 consonants.
So we have to select 2 vowels from the given 4 vowels.
The no. of ways we can select 2 vowels from the given 4 vowels is given by:
$ \Rightarrow {}^4{c_2} = 6$
We have to select 3 consonants from the given 5 consonants.
The no. of ways we can select 3 consonants from the given 5 consonants is given by:
$ \Rightarrow {}^5{c_3} = 10$
So now we have 2 vowels and 3 consonants, which means that we have 5 letters in total.
The no. of ways in which these 5 letters can be arranged is given by:
$ \Rightarrow 5! = 120$
Now in these 5 letters, the 2 vowels can be arranged in ${}^4{c_2}$ ways and the 3 consonants can be arranged in ${}^5{c_3}$ ways, which is given by:
$ \Rightarrow 120 \times {}^4{c_2} \times {}^5{c_3} = 120 \times 6 \times 10$
$ \Rightarrow 7200$

Final Answer: Total no. of words formed by using 2 vowels and 3 consonants taken from 4 vowels and 5 constants in equal to 7200.

Note:
While solving this problem please note that while considering the total no. of words which can be formed with using 2 vowels and 3 consonants taken from 4 vowels and 5 constants, here though the selected 5 letters can be arranged in 120 ways, but again in these 5 letters again the 2 vowels from the selected 4 vowels can be arranged in 6 ways and the 3 consonants from the selected 5 consonants can be arranged in 10 ways, hence multiplied with these.