Question

Out of 7 consonants and 4 vowels, the number of words (not necessarily meaningful) that can be made, each consisting of 3 consonants and 2 vowels, is
$\left( a \right)$ 24800
$\left( b \right)$ 25100
$\left( c \right)$ 25200
$\left( d \right)$ 25400

Answer 1

Hint: In this particular question use the concept that the number of ways to select r objects out of n is ${}^n{C_r}$ and the number of arrangements of these r objects is r!, so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
There are 7 consonants and 4 vowels.
Now we have made a five letter word with or without meaning each consisting 3 consonants and 2 vowels.
Now as we know that the number of ways to select r objects out of n is ${}^n{C_r}$
So the number of ways to select 3 consonants out of 7 is = ${}^7{C_3}$
And the number of ways to select 2 vowels out of 4 vowels is = ${}^4{C_2}$
So 5 letters are chosen now we have to arrange them.
As we all know that the number of ways to arrange n different objects are n!.
So the number of ways to arrange 5 different letters are 5!
So the total number of 5 letter words consisting of 3 consonants and 2 vowels are the multiplication of the above values so we have,
So the total number of 5 letter words consisting of 3 consonants and 2 vowels are = ${}^7{C_3} \times {}^4{C_2} \times 5!$
Now as we know that, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},n! = n\left( {n - 1} \right)\left( {n - 2} \right)......$ so using this property we have,
$ \Rightarrow {}^7{C_3} \times {}^4{C_2} \times 5! = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \times \left( {5.4.3.2.1} \right)$
Now simplify we have,
$ \Rightarrow {}^7{C_3} \times {}^4{C_2} \times 5! = \dfrac{{7!}}{{3!\left( 4 \right)!}} \times \dfrac{{4!}}{{2!\left( 2 \right)!}} \times \left( {120} \right)$
$ \Rightarrow {}^7{C_3} \times {}^4{C_2} \times 5! = \dfrac{{7.6.5.4!}}{{3!\left( 4 \right)!}} \times \dfrac{{4.3.2!}}{{2!\left( 2 \right)!}} \times \left( {120} \right)$
$ \Rightarrow {}^7{C_3} \times {}^4{C_2} \times 5! = \dfrac{{7.6.5}}{{3.2.1}} \times \dfrac{{4.3}}{{2.1}} \times \left( {120} \right)$
$ \Rightarrow {}^7{C_3} \times {}^4{C_2} \times 5! = 35 \times 6 \times \left( {120} \right) = 25200$
So this is the required answer.

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the combination as well as of the factorial which is given as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},n! = n\left( {n - 1} \right)\left( {n - 2} \right)......$, so simplify according to these formulas as above we will get the required answer.