Often we come across probelms in Permutations and combinations in which we are given a condition that two or more entities must be together, or that they cannot be together. We will try to look into those cases here.
Situation 1: 10 people are seated in a row, but A and B must be together, so in how many ways can we arrange them in a row.
In this case we tie a knot to A and B together, and treat them as 1 person first.
Next, we see the total number of people left, in this case those are 8.
So the total number of entities are 9.
Now these 9 entities can be seated in 9! ways.
But A and B, can be seated as AB or BA, hence we multiply the above number with 2!.
And our answer to the above problem is 9!*2!
Now we move on to "Gaps"
Situation 2: 10 people are seated in a row, but A and B cannot be together.
Now we know that A and B cannot be together, so let's talk about the other people first.
Those people can take 8 seats in 8! ways.
When those people take 8 seats gaps are created as follows:
Gap  G
People N
G N3 G N4 G N5 G N6 G N7 G N8 G N9 G N10 G
Now when you count the number of gaps among the 8 remaining people, excluding A and B, there are 9 gaps
A and B can take place in 9 of those gaps.
We can select those gaps in 9C2 ways.
And those two people can also be seated in 2! ways.
So the answer to situation 2 is
8! * 9C2 * 2!
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