Question

How many four letter words, with or without meaning, can be formed by using the letters of the word “EQUATION” such that at least one vowel should be included in each word and repetition of letters is not allowed?
A). 1680
B). 3410
C). 450
D). 2360

Answer 1

Hint: In the word “EQUATION”, there are 5 vowels and 3 consonants. To make a four letter word we can select one vowel and 3 consonants or 2 vowels and 2 consonants or 3 vowels and 1 consonant or only 4 vowels. Since the arrangement of the letters matters, use a permutation formula.

Complete step-by-step answer:
We know that a permutation is the choice or r thing from a set of n things without replacement and where the order matters.
The formula of permutation is:
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Here the given word is “EQUATION”. There are 8 letters in this word. 5 of them are vowels and 3 of them are consonants. We need to form 4 letter words such that at least one vowel should be there in each word.
That means, we can select 1 vowel and 3 consonants or 2 vowels and 2 consonants or 3 vowels and 1 consonant or 4 vowels.
If we change the order of selection then the word will also change. Therefore here the order of selection matters. Hence, we need to use the permutation formula.
Now, if we make the words using 1 vowel and 3 consonants, number of such words will be:
${}^{5}{{P}_{1}}\times {}^{3}{{P}_{3}}$
$=\dfrac{5!}{\left( 5-1 \right)!}\times \dfrac{3!}{\left( 3-3 \right)!}$
$=\dfrac{5!\times 3!}{4!\times 1}$ , as 0! = 1.
$=\dfrac{5\times 4!\times 6}{4!}=30$
If we make the words using 2 vowels and 2 consonants, number of such words will be:
${}^{5}{{P}_{2}}\times {}^{3}{{P}_{2}}$
$=\dfrac{5!}{\left( 5-2 \right)!}\times \dfrac{3!}{\left( 3-2 \right)!}$
$=\dfrac{5!\times 3!}{3!\times 1!}$
$=5!=5\times 4\times 3\times 2\times 1=120$
Now, if we make the words using 3 vowels and 1 consonant, number of such words will be:
${}^{5}{{P}_{3}}\times {}^{3}{{P}_{1}}$
$=\dfrac{5!}{\left( 5-3 \right)!}\times \dfrac{3!}{\left( 3-1 \right)!}$
$=\dfrac{5!\times 3!}{2!\times 2!}$
$=\dfrac{5\times 4\times 3\times 2!\times 3\times 2!}{2!\times 2!}=180$
Now, if we make the words using 4 vowels, number of such words will be:
${}^{5}{{P}_{4}}$
$=\dfrac{5!}{\left( 5-4 \right)!}$
$=\dfrac{5!}{1!}$
$=5!=5\times 4\times 3\times 2\times 1=120$
Therefore, total number 4 letter words with at least one vowel will be (30 + 120 + 180 + 120) = 450.
Hence, option (C) is correct.

Note: We can make mistakes while we are deciding to use permutation. We need to understand, if the arrangement of the letters changes, the word will also change. So the arrangement matters here. Therefore, we have to use permutation.