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 ${}^{n}{{P}_{r}}$ . 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 $=6$
To find the 4 letters word we have 4 vacant places. So, we can arrange them in the form of ${}^{6}{{P}_{4}}$ .
We know that –
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Here, we get –
$\begin{align}
& {}^{6}{{P}_{4}}=\dfrac{6!}{\left( 6-4 \right)!} \\
& \Rightarrow \dfrac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1} \\
\end{align}$
By cancelling the common factors from numerator and denominator, we get –
$\begin{align}
& =6\times 5\times 4\times 3 \\
& =360\text{ ways} \\
\end{align}$
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 ${}^{5}{{P}_{3}}$ , we get –
$\begin{align}
& {}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!} \\
& =\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \\
\end{align}$
By cancelling common factor from numerator & denominator we get –
$\begin{align}
& =5\times 4\times 3 \\
& =60\text{ ways} \\
\end{align}$
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 ${}^{5}{{P}_{3}}$ , we get –
$\begin{align}
& {}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!} \\
& =\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \\
\end{align}$
By cancelling common factor from numerator & denominator we get –
$\begin{align}
& =5\times 4\times 3 \\
& =60\text{ ways} \\
\end{align}$
Therefore, all possible No’s of arrangement
$\begin{align}
& =60\text{ ways + 60 ways} \\
& \text{=120 ways} \\
\end{align}$
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, $n=6.$
To find the 4 letter word we have 4 vacant places. i.e. $n!.$
Where, $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ...........$
No’s of filling of ${{1}^{st}}$ place $n=6$
No’s of filling of remaining in ${{2}^{nd}}$ place is $\left( n-1 \right)=5$
No’s of filling of remaining in ${{3}^{rd}}$ place is $\left( n-2 \right)=4$
No’s of filling of last place is $\left( n-3 \right)=3$
Therefore, $n!=6\times 5\times 4\times 3$
$=360\text{ ways}$
