Question

The different number of ways the letters of the word ‘YOGESH’ can be placed in the $8$ boxes of the given figure so that no row remains empty is

A.$26 \times 7!$

B.$26 \times 6!$

C.$26 \times 5!$

D.$26 \times 4!$

Answer 1

Hint: The total number of ways in which rows can be filled is $\left( \begin{array}{l}
8\\
6
\end{array} \right)$ ways. Out of these there will be two cases in which the row is empty. So the ways in which no row is empty is $\left( \begin{array}{l}
8\\
6
\end{array} \right) - 2$. We have to solve in how many ways does the word ‘YOGESH’ can be arranged and multiply with $\left( \begin{array}{l}
8\\
6
\end{array} \right) - 2$.



Complete step by step solution
Given:

The number of boxes is $8$.

It is given that the word is ‘YOGESH’.

The total number of ways in the rows can be filled is $\left( \begin{array}{l}
8\\
6
\end{array} \right)$ ways.

We know the formula for $\left( \begin{array}{l}
n\\
r
\end{array} \right)$ is given by,
 $\left( \begin{array}{l}
n\\
r
\end{array} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

On substituting the values in the above expression, we get,
 $\left( \begin{array}{l}
8\\
6
\end{array} \right) = \dfrac{{8!}}{{\left( {8 - 6} \right)!6!}}$

The formula we can use for $n!$ is given as,
 $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ... \times 2 \times 1$

On applying the formula for $n!$ in the above expression $\left( \begin{array}{l}
8\\
6
\end{array} \right) = \dfrac{{8!}}{{\left( {8 - 6} \right)!6!}}$we get,
$\begin{array}{c}
\dfrac{{8 \times 7}}{{2 \times 1}} = 4 \times 7\\
 = 28
\end{array}$

The total number of ways for the rows to be filled is $28$ways.

Out of these there will be two cases in which a row will be empty. That is the case where the top row is empty and the case where the bottom row is empty.

Hence the ways in which any of the row is empty is $28 - 2$ ways which is $26$ways.

Now, In ‘YOGESH’ there are 6 words and The number of ways the ‘YOGESH’ is arranged is,
$6!$

So, the total number of ways of placing in $8$ box is,
 $26 \times 6!$

Hence, the correct option is (B) and $26 \times 6!$ is the correct answer.

Note: In the ‘YOGESH’ there is no common letter so only the number of ways are $6!$ways. For example, if we take ‘KEEP’ in this word there are two common letters hence the arrangements will be $\dfrac{{4!}}{{2!}}$.