Question

In how many ways can 4 consonants and 2 vowels be selected in the English alphabet consisting of 21 consonants and 5 vowels?

Answer 1

We first separate the groups in which the consonants have the majority. We separately find the number of ways we can choose 4 consonants and 2 vowels from 21 consonants and 5 vowels. The general form of combination is ${}^{n}{{C}_{r}}$. It’s used to express the notion of choosing r objects out of n objects. We multiply them to find the solution.

Complete step-by-step solution:
There are in total 21 consonants and 5 vowels out of which we need to select 4 consonants and 2 vowels. The notion of choosing r objects out of n objects is denoted by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$.
The number of choices for 4 consonants out of 21 consonants will be ${}^{21}{{C}_{4}}=\dfrac{21!}{4!\times 17!}=5985$ ways.
The number of choices for 2 vowels out of 5 vowels will be ${}^{5}{{C}_{2}}=\dfrac{5!}{2!\times 3!}=10$ ways.
Total will be $5985\times 10=59850$.
Therefore, the number of ways 4 consonants and 2 vowels can be selected in the English alphabet consisting of 21 consonants and 5 vowels is 59850.

Note: There are some constraints in the form of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. There is no need for permutation of the selected alphabets. The problem is about choosing the alphabets only while permutation is used for arrangement of things.