Question

A word has 8 consonants and 3 vowels. How many distinct words can be formed if $4$ constant and $2$ vowels are chosen?

Answer 1

To solve the question you need to know the knowledge of combination and permutation. We need to calculate the combination for the selection of the consonants and vowels, using the formula which is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . The next step will be the arrangement of the letters which will be done using permutation, using formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Complete step-by-step solution:
The question asks us to find the number of distinct words that can be formed using $4$ constant and $2$ vowels if we are given with $8$ consonants and $3$ vowels.
The first step to solve the question will be to find the number of groups that can be formed using $4$ constant and $2$ vowels when we are given with $8$ consonants and $3$ vowels. Total number of groups formed will be calculated using combination, so the number of groups that can be formed are:
$\Rightarrow {}^{8}{{C}_{4}}\times {}^{3}{{C}_{2}}$
On calculating the combination given above we get:
$= \dfrac{8!}{4!\left( 8-4 \right)!}\times \dfrac{3!}{2!\left( 3-2 \right)!}$
$= \dfrac{8!}{4!\left( 4 \right)!}\times \dfrac{3!}{2!\left( 1 \right)!}$
Now we will be solving for the factorial of the numbers present in the expression above. The factorial of the number $n$ is represented as $n!$ which is the product of all the numbers from $1$ to $n$ . The formula which we will use here will be $n!=n\left( n-1 \right)!$ . So on applying the same we get:
$= \dfrac{8\times 7\times 6\times 5\times 4!}{(4\times 3\times 2\times 1)4!}\times \dfrac{3\times 2!}{2!\left( 1 \right)!}$
Cancelling out the terms common in numerator and in the denominator we get:
$= \dfrac{7\times 6\times 5}{1}\times \dfrac{1}{1}$
On final calculation we get:
$=210$
The number of groups of letters found to be $210$ . To find the total number of words formed we will multiply $210$ to the factorial number of letters as different arrangements of the letter will form different words. Here we will be using the concept of permutation as we are required to arrange the letters. So on calculating we get:
$= 210\times {}^{6}{{P}_{6}}$
$= 210\times 6!$
On calculating we get:
$= 210\times 6\times 5\times 4\times 3\times 2\times 1$
$= 151200$
$\therefore$ Total distinct words that can be formed using$4$ constant and $2$ vowels from has $8$ consonants and $3$ vowels is $151200$.
Note: For the selection purpose ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ is used, which actually determine the number of possible arrangements. Factorial of a number is the product of all the numbers from one to the number itself, which means $n!=n(n-1)(n-2)....2.1$. The formula which has played an important role in the above solution and is helpful is to expand $n!$ as the product of numbers and factorials. So $n!$ could be written in many ways like $n!=n(n-1)!$ , $n!=n(n-1)(n-2)!$ and so on, as per the demand of the problem.