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You are given 7 different consonants and 3 different vowels. You have to form three letter words each containing 2 consonants and 1 vowel so that the vowel is always in between the two consonants. How many such words can be formed?

Last updated date: 20th Sep 2024
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Hint: If we are required to select r things from n different things (n > r), then the number of ways in which we can select these are things is given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Also, if we are required to arrange these are things among themselves, the number of ways of arrangement is r!. Using this, we can solve this question.

Complete step-by-step solution -
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In permutations and combinations, if we are given n different things from which, we are required to select r things such that (n > r), then the number of ways in which we can do so is given by the formula,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . . . . . . . . . . . . . (1)
If we are required to arrange these ‘r’ things among themselves, the number of ways of arrangement is r! . . . . . . . . . (2)
In the question, we are given 7 different consonants and 3 different vowels. We are required to form three letter words each containing 2 consonants and 1 vowel so that the vowel is always in between the two consonants.
Using formula (1), the number of ways in which we can select 2 consonants from 7 different consonants is equal to,
\begin{align} & {}^{7}{{C}_{2}}=\dfrac{7!}{2!\left( 7-2 \right)!} \\ & \Rightarrow {}^{7}{{C}_{2}}=\dfrac{7\times 6\times 5!}{2\left( 5 \right)!} \\ & \Rightarrow {}^{7}{{C}_{2}}=21 \\ \end{align}
Using formula (1), the number of ways in which we can select 1 vowel from 3 different vowels is equal to,
\begin{align} & {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!} \\ & \Rightarrow {}^{1}{{C}_{1}}=\dfrac{3\times 2!}{\left( 2 \right)!} \\ & \Rightarrow {}^{3}{{C}_{1}}=3 \\ \end{align}
So, the number of words containing 2 consonants and 1 vowel is given by multiplying the above two obtained number is equal to,
$21\times 3=63$
Now in these 63 words, we have to arrange the vowels and consonants in such a way that the consonants hold the first and the third place and the vowel holds the middle i.e. second place. Using formula (2), the number of ways in which the 2 consonants can be arranged within themselves is equal to 2! = 2.
Hence, the total number of words in which the vowel is always in between the two consonants is equal to $63\times 2=126$.

Note: There is a possibility that one may forget to arrange the two consonants after their selection which will lead us to an incorrect answer. Each arrangement of the two consonants will give us different words. That is why there is a need to arrange them within themselves.