Question

How many three-letter words can be formed by using the letters of the word ‘EQUATION’ such that at least one vowel should be included in each word? (Repetition of letters not allowed)
(a) 336
(b) 6
(c) 339
(d) 330

Answer 1

Hint: First of all, consider the word EQUATION and separate consonant and variable. Now, take 3 cases and use them to find the number of words. Case 1 is when there would be 1 vowel and 2 consonants, case 2 would have 2 vowels and 1 consonant and case 3 would have 3 vowels and 0 consonants.

Complete step-by-step solution -
In this question, we have to find the number of three-letter words that can be formed by using the letters of the word ‘EQUATION’ such that at least one vowel should be included in each word. Also, the repetition of letters is not allowed.
Let us consider the word given in the question
$\underline{\text{E}}\text{ }\underline{\text{Q}}\text{ }\underline{\text{U}}\text{ }\underline{\text{A}}\text{ }\underline{\text{T}}\text{ }\underline{\text{I}}\text{ }\underline{\text{O}}\text{ }\underline{\text{N}}$
In the above word, we can see that there are five distinct vowels and that are E, U, A, I, O, and the remaining letters are Q, T, and N.
We know that multiplication Theorem of probability states that If A and B are two independent events, then the probability that both will occur is equal to the product of their individual probabilities.
Here we can see that selection of vowels, selection of consonants, and arrangements of these letters are three independent events, that is, their probability of occurrence is not affected by each other. So, the probability that these three events occur simultaneously is equal to the product of their individual probabilities.
Now, let us make three-letter words from these letters with at least one vowel. Here, there would be three cases as follows:
1. We can make three-letter words with 1 vowel and 2 consonants. So, by using multiplication theorem, we get,
The number of words = (Number of the selection of 1 vowel out of 5 vowels) $\times$ (Number of the selection of 2 consonants out of 3 consonants) $\times$ (Number of the arrangement of these three letters)
We know that selection of r out of n things is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and the arrangement of m things is given by m! By using these, we get,
$={{\text{ }}^{5}}{{C}_{1}}\times {{\text{ }}^{3}}{{C}_{2}}\times 3!$
$=\dfrac{5!}{1!4!}\times \dfrac{3!}{2!1!}\times 3!$
$=\dfrac{\left( 5\times 4! \right)}{4!}\times \dfrac{\left( 3\times 2! \right)}{2!}\times 3!$
$=5\times 3\times 3\times 2\times 1$
= 90 words
2. We can make three-letter words with 2 vowels and 1 consonant. So, by using multiplication theorem, we get,
The number of words = (Number of the selection of 2 vowels out of 5 vowels) $\times$ (Number of the selection of 1 consonant out of 3 consonants) $\times$ (Number of the arrangement of these three letters)
We know that selection of r out of n things is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and the arrangement of m things is given by m! By using these, we get,
$={{\text{ }}^{5}}{{C}_{2}}\times {{\text{ }}^{3}}{{C}_{1}}\times 3!$
$=\dfrac{5!}{2!3!}\times \dfrac{3!}{1!2!}\times 3!$
$=\dfrac{\left( 5\times 4\times 3! \right)}{2!3!}\times \dfrac{\left( 3\times 2! \right)}{2!}\times 3!$
$=\dfrac{20}{2}\times 3\times 3\times 2$
= 180 words
3. We can make three-letter words with 3 vowels and 0 consonants. So, by using multiplication theorem, we get,
Number of words = (Number of the selection of 3 vowels out of 5 vowels) $\times$ (Number of the arrangement of these 3 letters)
We know that selection of r out of n things is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and the arrangement of m things is given by m! By using these, we get,
$={{\text{ }}^{5}}{{C}_{3}}\times 3!$
$=\dfrac{5!}{3!2!}\times 3!$
$=\dfrac{5!}{2!}$
$=\dfrac{5\times 4\times 3\times 2!}{2!}$
= 60 words
So, we get the total words = 90 + 180 + 60 = 330 words.
So, option (d) is the right answer.

Note: Students can also solve the questions involving the word ‘at least’ by using (Total) – (None) method as follows:
Number of 3 letter words such that at least one letter is vowel = Total 3 letter words – words containing no vowels
We know that from the letters of words EQUATION
Total 3 letter words without repetition $=\underline{8}\times \underline{7}\times \underline{6}=336$
Total 3 letter words containing no vowels with repetition $=\underline{3}\times \underline{2}\times \underline{1}=6$
So, we get the number of 3 letter words such that at least one letter is vowel = 336 – 6 = 330 vowels.