Question

Choose the correct option provided below for the following question.
How many different words can be formed from $12$ consonants and $5$ vowels by taking $4$ consonants and $3$ vowels in each word?
A) $4950 \times 7!$
B) $7{!^2}$
C) ${}^{12}{C_4} \times {}^5{C_3}$
D) None of these

Answer 1

We notice the given total number of objects and the number of objects to be chosen. We take the combinations of the respective conditions then multiply them to combine the overall combination. We simplify the factorials until it is in the simplest form to get the final answer.

Complete step-by-step solution:
Here it is given that,
Given number of consonants $= 12$
Given number of vowels $= 5$
Number of consonants taken $= 4$
Number of vowels taken $= 3$
Given that $4$ consonants are taken from $12$ consonants.
\Rightarrow $$$$4 consonants out of $12$ can be selected in ${}^{12}{C_4}$ ways.
Given that $3$ vowels are taken from $5$ vowels.
\Rightarrow $$$$3 vowels out of $5$ can be selected in ${}^5{C_3}$ ways.
In each group we can write the letters in $\left( {3 + 4} \right) = 7$ different ways.
$\Rightarrow$ We can have $7!$ ways to write a word in the given combinations.
Number of words that can be formed using the above combinations;
${ \Rightarrow ^{12}}{C_4} \times {}^5{C_3} \times 7!$
Expanding the combination terms, we get,
\Rightarrow $$$$\dfrac{{12!}}{{\left( {12 - 4} \right)! \cdot 4!}} \times \dfrac{{5!}}{{\left( {5 - 3} \right)! \cdot 3!}} \times 7!
Simplifying the above terms, we get:
\Rightarrow $$$$\dfrac{{12!}}{{8!4!}} \times \dfrac{{5!}}{{2!3!}} \times 7!
Furthermore, expanding the factorials, that is expanding $12!$ and $5!$ in terms of cancelling the denominator, simplifying it, we get:
$\Rightarrow \dfrac{{12 \times 11 \times 10 \times 9 \times 8!}}{{8!4!}} \times \dfrac{{5 \times 4 \times 3!}}{{2!3!}} \times 7!$
Cancelling out the same terms in the numerator and denominator, we get:
$\Rightarrow \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{5 \times 4}}{{2 \times 1}} \times 7!$
Cancelling out the terms to reduce it into the simplest form;
$\Rightarrow 11 \times 5 \times 9 \times 5 \times 2 \times 7!$
Multiplying all the terms, we get;
$4950 \times 7!$
Hence, the number of different words that can be formed from $12$ consonants and $5$ vowels by taking $4$ consonants and $3$ vowels in each word are $4950 \times 7!$

$\therefore$ The correct option is A.

Note: We have to know the formula for combination and permutation must be differentiated clearly. Here, the binomial theorem of combinations is used to find the final solution. According to the binomial theorem of combinations, if there are $n$ objects from which $r$ objects are chosen and formed in a combination, it is represented as ${}^n{C_r}$.
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$