Question

Consider all the permutations of the word \[''JANLOKPAL''\]. Number of words in which O never lies between \[J\] & \[N\] is equal to -
\[(A)6\left| \!{\underline {\,
  7 \,}} \right. \]
\[(B)8\left| \!{\underline {\,
  7 \,}} \right. \]
\[(C)9\left| \!{\underline {\,
  7 \,}} \right. \]
\[(D)12\left| \!{\underline {\,
  7 \,}} \right. \]

Answer 1

Hint: 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 \[=(28+35+40+43+44+28+35+40+43)\times \dfrac{6!}{2!2!}\]
 \[=(2(28+35+40+43)+44)\times \dfrac{6!}{2!2!}\]
\[\begin{align}
  & =336\times \dfrac{6!}{2!2!} \\
 & \\
\end{align}\]
 \[\begin{align}
  & =12\times 4\times 7\times \dfrac{6!}{2\times 2} \\
 & =12\times 7! \\
\end{align}\]
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.