Answer

Verified

446.1k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE