the number of different ways in which 8 persons can stand in a row so that between 2 particular person A and B there are always 2 persons​a. 60*5!​b. 15*51*5!c. 4!*5!d. 15*5!*4!

Daniel
15 Points
7 years ago
We will select two persons from 6 others by 6c2 then make a string of the selected person and (AB) then will arrange the string along with 4 rest people by 5! And then arrange people in the string by2!*2! Giving- (6c2*2!*2!)(5!)=(15*4)(120)=3600...hope this helps.
Akash
13 Points
4 years ago
Here we are told that “two people in particular, A and B”... So we can’t make selection of 2 from 6. Now we name the positions from 1 to 8.  So now consider A & B occupy position 1 and 4. They can occupy these 2 positions in 2 ways(2P2). Rest 6 positions can be filled by remaining 6 people in 6! ( Six Factorial) ways. That’s means if A and B were to occupy portion 1&4, we have 2*(6!) ways of arranging them. Now A&B can occupy 2&5, 3&6, 4&7, 5&8... thus total of 5 cases and hence total permutations are 5*2*(6!)= 60*(5!)...i.e. Option (A)... Book R.D. Sharma says the same... not 3600.
Lipin
13 Points
4 years ago
First select 2 person from 6(the rest will be A and B)
Number of arranging 2 from 6 =6C2=15
Condition ====A xx B
So remaining 4 people including these can be arranged in 5!
And they can be placed in 4 ways
Total number of ways=15 * 5!*4
=7200=60*5!