Question

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
A. 50000
B. 50100
C. 50300
D. 50400

Answer 1

Hint: Selection of r things out of ‘n’ different thing is given by relation $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$ and arrangement of n different things is given by relation n! = 1.2.3.4………n. Select 2 – 2 letters from consonants and variables and hence arrange them to get the total number of 4 letter words.
Complete step by step answer:
As it is given that English alphabet has 5 vowels and 21 consonants and hence, we need to form words with the two different vowels and 2 different consonants. It means we need to form 4 letter words with all letters different.

 4 letter word with 2 different consonants and 2 different vowels.
As we know that there are 21 consonants out of which only 2 consonants will be used for formation of the word and similarly, there are 5 vowels and we need to select 2 vowels of them.
So, we have to select 2 consonants and 2 vowels from 21 consonants and 5 vowels respectively.
 As we know the number of ways of selection of ‘r’ things out of ‘n’ things is given as $C_{r}^{n}$ whose formula is given as
$C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$
 Where n! = 1.2.3.4………n.
Hence we can select 2 vowels from 5 vowels by$C_{2}^{5}$ ways and select 2 consonants from 2 consonants from 21 consonants by $C_{2}^{21}$ways.
Now, we have selected 4 letters to form words of 4 letters. Hence, now we can arrange these 4 different 4 letters to get the number of words possible from that selected 4 letters.
As we know, the total number of arrangements of n different things can be given by n!
So, after selecting 4 letters, arrangement of them can be done by 4! Ways. Hence number of words of 4 different letters by selecting 2 consonants and 2 vowels from 21 consonants and 5 vowels can be given by
$\Rightarrow C_{2}^{5}\times C_{2}^{21}\times 4!$
Now, use the identity
$C_{r}^{n}=\dfrac{n!}{\left( n-r \right)!r!}$ to solve $C_{2}^{5},C_{2}^{21}$
So, we get Total number of words
 $\begin{align}
  & =\dfrac{5!}{2!3!}\times \dfrac{21!}{2!19!}\times 4! \\
 & =\dfrac{5\times 4}{2}\times \dfrac{21\times 20}{2}\times 4! \\
 & =10\times 210\times 24 \\
\end{align}$
Total number of words = 50400
Hence, option (d) is the correct answer.
Note: One may calculate the arrangement by the identity $P_{r}^{n}$ directly and can give answer as $P_{2}^{21}\times P_{2}^{5}$, which is wrong as here, we are arranging only two letters of consonants and two letters of vowels but we need to arrange all the four letters to get the total number of words. With using the above formula, we will miss a lot of words as well. So, take care of it and be careful.
Do selection and then look for selection with these types of questions, don’t try to do both at the same time otherwise, we may get wrong results as well.
Don’t calculate 10!, 6!, 4! Individually.
Use the identity given as:
n! = n.(n -1)(n – 2)……….r!
Where $r\le n$
Example: 10! = 10.9.8.7! = 10.9.8.7.6.5! = 10.9!