Question

The number of different ways in which 8 persons can stand in a row so that between two particular person A and B there is always two persons, is
A) $60(5!)$
B) $15(4!)(5!)$
C) $4!5!$
D) None of these

Answer 1

Hint:
Here, we have to arrange 8 persons in different ways in which 2 persons are always between the person A and B. So firstly, we have to arrange 2 people between the A and B and then find the ways in which other people are arranged.

Complete step by step solution:
Here, firstly we have to choose a two person in between A and B.
The two people can be chosen in ${C_2}^{6}$ ways.
Now, we will take the above four persons as a one group.
Therefore, the one group and the remaining four persons can be arranged in $5!$ ways.
Also A and B can be arranged in 2 ways each.
The total ways in the one group arranged $ = 2 \times 2$
Therefore, the total number of arrangement possible is equal to
 $
   = {}^6{C_2} \times 5! \times 4 \\
   = \dfrac{{6!}}{{(6 - 2)! \times 2!}} \times 5! \times 4 \\
   = \dfrac{{30}}{2} \times 5! \times 4 \\
   = 15 \times 4 \times 5! \\
   = 60 \times 5! \\
 $

Therefore, the 8 persons in which two persons are between A and B can be arranged in $60 \times 5!$

Note:
When the events are dependent on each other we multiply all the events and when the events are independent of each other we add all the events. Permutation is applicable when the order of selection is important and combination is applicable when the order of the selection is not important.