Question

How many five letter words containing 3 vowels and 2 consonants can be formed using the letters of the word ‘EQUATION’ so that the two consonants occur together?

Answer 1

Hint: Here we need to find the number of different five letter words that can be formed using the letters in the given word. For that, we will first count the number of letters present in the given word and then we will count the number of vowels and number of consonants present in the word. Then we will find the number of ways to arrange these using the formula of permutation. After simplification, we will get the required value.

Complete step-by-step answer:
The given word is ‘EQUATION’.
Total number of letters available in the word ‘EQUATION’ is equal to 8.
We can see from the word that there are five vowels in the word ‘EQUATION’ i.e. U, A, I, O, E and there are three consonants in the word ‘EQUATION’ i.e. Q, T, N.
We need to find the total number of words of five letters containing 3 vowels and 2 consonants. Therefore,
Number of ways to select three vowels $$={}^5{P}_3$$
Using this formula $${}^n{C}_r={\displaystyle \frac{n!}{\left(n-r\right)!r!}}$$, we get
 Number of ways to select three vowels $$={\displaystyle \frac{5!}{\left(5-3\right)!\times 3!}}$$
On further simplification, we get
Number of ways to select three vowels $$={\displaystyle \frac{5!}{2!\times 3!}}$$
Computing the factorial, we get
Number of ways to select three vowels $$={\displaystyle \frac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}}=5\times 2=10$$
For consonants, we have
Number of ways to select two consonants $$={}^3{C}_2$$
Using this formula $${}^n{C}_r={\displaystyle \frac{n!}{\left(n-r\right)!r!}}$$, we get
Number of ways to select two consonants $$={\displaystyle \frac{3!}{\left(3-2\right)!\times 2!}}$$
On further simplification, we get
Number of ways to select two consonants $$={\displaystyle \frac{3!}{1!\times 2!}}$$
Computing the factorial, we get
Number of ways to select two consonants $$={\displaystyle \frac{3\times 2\times 1}{1\times 2\times 1}}=3$$
Since, two consonants always occur together, so we will consider two consonants as one.
Number of ways to arrange these 5 letters considering two consonants as one $$=4!\times 2!$$
On further simplification, we get
Number of ways to arrange these 5 letters considering two consonants as one $$=4\times 3\times 2\times 1\times 2\times 1=48$$
Hence, the number of different words formed by these 5 letters $$=10\times 3\times 48=1440$$
Therefore, the correct answer is equal to 1440.

Note: Here we have used the formula of combination and not permutation to get the required number of ways. Therefore we need to know the basic difference between the permutation and combination to avoid any mistakes. Permutation is used when we have to find the possible elements but the combination is used when we need to find the number of ways to select a number from the collection.