Question

in how many ways can 6 boys be arranged in a line so that 3 particular boys may never come together?

Answer 1

they can be first third and fifth or second fourth and sixth, and there are \(3!\) choice for each , and then \(3!\) choices for the other three slots so \(2\times 3!\times 3!\)

Answer 2

@kropot72 look reasonable? just off the top of my head

Answer 3

@satellite73 Looks exactly correct! Very good reasoning :)

Answer 4

But I did not get some part :(

Why you multiplied it with 2?? @satellite73

Answer 5

When there are no restrictions the 6 boys can be arranged in a row in 6! = 720 way
When the 3 boys are treated as one "unit", there are 4 persons and these
4 persons can be arranged in 4 ! = 24 ways
The unit boys can be arranged among themselves in 3 ! = 6 ways.

The number of ways the 3 boys sit together is 4! × 3!= 144

Now, there are only 2 possibilities: the boys are together or they are not.

So the number of ways of arranging them so that the 3 boys are NOT together is:
720 - 144 = 576

Answer 6

So who is right??

Answer 7

i once got a similar problem, and the answer was also biggere tham i thought.

Answer 8

There are two ways of interpreting this question:
1. The three boys that may never come together cannot come together as a trio and cannot come together as a pair.
2. The three boys that may never come together cannot come together as a trio but can come together as a pair.
The solution by @satellite73 follows interpretation #1 and the solution by @Luis_Rivera follows interpretation #2.