Asmathumairy wrote:Hi,
May any one help me with this problem with a clear explanation?
I do not know how to deal with it.


six children will sit in a row of six chairs, but Jackie and Marilyn cannot be seated next to each other.

How many arrangements are possible?

Thanks in advance.

Start by determining the number of possible seating arrangements we'd have if there were no restrictions. We have 6 children and 6 chairs, so there's 6! = 720 ways the children can be seated. (Any of the 6 children could sit in the first seat, any of the remaining 5 can sit in the next one, and so forth.)

Now we want to subtract out the undesired outcomes. The undesired scenario is when Jackie and Marilyn sit next to each other. So imagine that our six children are A, B, C, D, J, and M. Let's force J and M to sit next to each other by fusing them together at the hip. Now, for all intents and purposes, we have 5 'children.' A, B, C, D, and J-M. If we have 5 children, there's be 5!, or 120 undesirable ways we could sit them. But in this scenario, we have J to the left of M. They could also switch positions, so that our 5 children would be A, B, C, D, and M-J. There's be another 5!, or 120 undesirable ways we can seat these children. In all, there's be 120 + 120 = 240 undesired scenarios in which J and M are next to each other.

Last: subtract the undesired from the total: 720 - 240 =480