Question

Consider all the permutations of the word $''JANLOKPAL''$. Number of words in which O never lies between $J$ & $N$ is equal to -

7 \,}} \right. $$ $$(B)8\left| \\!{\underline {\, 7 \,}} \right. $$ $$(C)9\left| \\!{\underline {\, 7 \,}} \right. $$ $$(D)12\left| \\!{\underline {\, 7 \,}} \right. $$

Answer 1

In this type of question you can think of this way that here are occurring some different cases for distinct initial positions of $J$ and $N$, and for each case you can calculate the permutation of rest letters keeping in mind that $O$ cannot take place between $J$ and $N$.

Complete answer:
We have the word $''JANLOKPAL''$ which have total $9$ letters in which some are occurring more than once and that are$2A's$,$2L's$and $5$ are unique.
Now we will make different cases for $J$ and $N$ and permute all the rest of the letters and we have to permute letters in each case so that $O$ cannot come between $J$ and $N$.
Now take following Cases ,
Case $1$:
In this case we are setting $'J'$at first position out of nine positions, and we will count permutation for all possible different values for$'N'$.
So, first we put $'N'$ at position two that is just next to$'J'$.
So, we have $7$positions left for $'O'$ out of $9$positions.
So number of words form in this case$=7\times \dfrac{6!}{2!2!}$
Since after setting positions for $3$letters we have left $6$letters having $2$letters $2$reoccurrences, therefore to nullify repeated permutation we divide it by $2!2!$.
So now, we put $'N'$ at position three that is $'J'$ and $'N'$ has now $1$ space in between.
Now since $'O'$cannot take that in between position so choices left for $'O'$ is $6$but for rest letters choices remains same.
Therefore, number of words form in this case$=6\times \dfrac{6!}{2!2!}$
Now, we can think logically a few more steps in Case $1$as we move the position of $'N'$ rightmost choices for $'O'$ will decrease.
So, we will find pattern in this case and that will be:
$7\times \dfrac{6!}{2!2!}+6\times \dfrac{6!}{2!2!}+5\times \dfrac{6!}{2!2!}+.......+1\times \dfrac{6!}{2!2!}$
Therefore total number of words formed in this case
$=(7+6+5+4+3+2+1)\times \dfrac{6!}{2!2!}$
$=(28)\times \dfrac{6!}{2!2!}$
Similarly, Case $2$
Now put $'J'$ at second position.
So, in this total number of words formed in this case
$=(7+6+5+4+3+2+1+7)\times \dfrac{6!}{2!2!}$
$=(35)\times \dfrac{6!}{2!2!}$
Similarly, Case 3
Total number of words in this case
$=(7+6+7+6+5+4+3+2)\times \dfrac{6!}{2!2!}$
$=(40)\times \dfrac{6!}{2!2!}$
Now as the pattern is very clear, using this pattern only now will we find the value in all the cases.
In case $4$, total number of words
$=(7+6+5+7+6+5+4+3)\times \dfrac{6!}{2!2!}$
$=(43)\times \dfrac{6!}{2!2!}$
In Case $5$, total number of words
$=(7+6+5+4+7+6+5+4)\times \dfrac{6!}{2!2!}$
$=(44)\times \dfrac{6!}{2!2!}$
In case $6$, total number of words
$=(7+6+5+7+6+5+4+3)\times \dfrac{6!}{2!2!}$
$=(43)\times \dfrac{6!}{2!2!}$
In case $7$, total number of words
$=(7+6+7+6+5+4+3+2)\times \dfrac{6!}{2!2!}$
$=(40)\times \dfrac{6!}{2!2!}$
In case $8$, total number of words
$=(7+6+5+4+3+2+1+7)\times \dfrac{6!}{2!2!}$
$=(35)\times \dfrac{6!}{2!2!}$
In case 9, total number of words
$=(7+6+5+4+3+2+1)\times \dfrac{6!}{2!2!}$
$=(28)\times \dfrac{6!}{2!2!}$
Therefore, total words $=(28+35+40+43+44+28+35+40+43)\times \dfrac{6!}{2!2!}$
$=(2(28+35+40+43)+44)\times \dfrac{6!}{2!2!}$

& =336\times \dfrac{6!}{2!2!} \\\ & \\\ \end{aligned}$$ $$\begin{aligned} & =12\times 4\times 7\times \dfrac{6!}{2\times 2} \\\ & =12\times 7! \\\ \end{aligned}$$ **Therefore, Option $$D$$ is correct.** **Note:** Permutation can be defined as the phenomenon of rearranging the data elements in all different possible order. Whereas, combination can be defined as a process of selecting items from a given collection. Permutation and Combination are interrelated to each other.