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Question: How many 4 letter words can be formed from the letters of the word ‘ANSWER’ ? How many of the words ...

How many 4 letter words can be formed from the letters of the word ‘ANSWER’ ? How many of the words start with a vowel?

Explanation

Solution

Hint: First we will find the 4 letter words from the letters of the word ‘ANSWER’ by arranging the total letters, i.e 6 in the form of nPr{}^{n}{{P}_{r}} . Then we will find the 4 letter words start with vowels for which two cases arise. The two vowels are A and E. After finding the number of ways for these cases, we have to add them for the final answer.

Complete step by step solution:
Here, first we will find the 4 letter words can be formed from the letters of the word ‘ANSWER’.
Total No’s of letters in the word ‘ANSWER’ is =6=6
To find the 4 letters word we have 4 vacant places. So, we can arrange them in the form of 6P4{}^{6}{{P}_{4}} .
We know that –
nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
Here, we get –
6P4=6!(6−4)! ⇒6×5×4×3×2×12×1 \begin{aligned} & {}^{6}{{P}_{4}}=\dfrac{6!}{\left( 6-4 \right)!} \\\ & \Rightarrow \dfrac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1} \\\ \end{aligned}
By cancelling the common factors from numerator and denominator, we get –
=6×5×4×3 =360 ways \begin{aligned} & =6\times 5\times 4\times 3 \\\ & =360\text{ ways} \\\ \end{aligned}
Now, we will find the No’s of 4 letter words start with vowels.
There are two vowels in the word ‘ANSWER’ i.e. ‘A’ & ‘E’, so here two cases arises:
Case 1: Words start with a letter A.
If the first letter be ‘A’ then the word be: A __ __ __.
Here, we left with 3 vacant places from the remaining 5 letters.
So, we can arrange them in 5P3{}^{5}{{P}_{3}} , we get –
5P3=5!(5−3)! =5×4×3×2×12×1 \begin{aligned} & {}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!} \\\ & =\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \\\ \end{aligned}
By cancelling common factor from numerator & denominator we get –
=5×4×3 =60 ways \begin{aligned} & =5\times 4\times 3 \\\ & =60\text{ ways} \\\ \end{aligned}
Case 2: Words start with a letter E.
If the first letter be ‘A’ then the word be: E __ __ __.
Here, we left with 3 vacant places from the remaining 5 letters.
So, we can arrange them in 5P3{}^{5}{{P}_{3}} , we get –
5P3=5!(5−3)! =5×4×3×2×12×1 \begin{aligned} & {}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!} \\\ & =\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \\\ \end{aligned}
By cancelling common factor from numerator & denominator we get –
=5×4×3 =60 ways \begin{aligned} & =5\times 4\times 3 \\\ & =60\text{ ways} \\\ \end{aligned}
Therefore, all possible No’s of arrangement
=60 ways + 60 ways =120 ways \begin{aligned} & =60\text{ ways + 60 ways} \\\ & \text{=120 ways} \\\ \end{aligned}
Hence, 360 ways of 4 letter words can be formed from the letters of the word ‘ANSWER’ and 120 ways of 4 letter words start with vowels.

Note: Students should solve this problem very carefully. They may make a mistake while taking the value of n as 4 (4 letter word) instead of 6.
Students can also solve this in an alternative way. Total No’s of letters in the word ‘ANSWER’ is 6.
So, n=6.n=6.
To find the 4 letter word we have 4 vacant places. i.e. n!.n!.
Where, n!=n×(n−1)×(n−2)×...........n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ...........
No’s of filling of 1st{{1}^{st}} place n=6n=6
No’s of filling of remaining in 2nd{{2}^{nd}} place is (n−1)=5\left( n-1 \right)=5
No’s of filling of remaining in 3rd{{3}^{rd}} place is (n−2)=4\left( n-2 \right)=4
No’s of filling of last place is (n−3)=3\left( n-3 \right)=3
Therefore, n!=6×5×4×3n!=6\times 5\times 4\times 3
=360 ways=360\text{ ways}