SOLUTION: 7 girls and 5 boys are to be seated around a table. Find the number of ways if 3 particular girls must be together, and 2 particular boys must not be adjacent?

Whenever we have a problem involving people sitting around a table,
we consider it as though the table were placed on a large turntable
and could be rotated clockwise or counter-clockwise with no change
in the seating arrangement.

1. First we'll enumerate the ways with the 3 girls sitting together
but with no restriction on the boys.
2. Then we'll enumerate the cases with the 3 girls together and the 2 boys
together.
3. Then we'll subtract the result of 2 from the result of 1.
 
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1. We enumerate the ways with the 3 girls sitting together
but with no restriction on the boys.

The 3 girls can be arranged together in 3! = 6 ways, 

So for each of those 6 ways that we can put the girls together, we can
seat the remaining 9 people around the 3 girls in 9! ways.

That's (3!)(9!) = (6)(362880) = 2177280 ways.

 
2. We enumerate the cases with the 3 girls together and the 2 boys
together.

As before, the 3 girls can be arranged together in 3! = 6 ways, 
The 2 boys can be arranged together in 2! = 2 ways.

So that's (3!)(2!) = (6)(2) = 12 ways to place the 3 girls together and 
the 2 boys together.

Now, instead of having 9 people to place around the 3 girls, we only have 8
"THINGS", 4 individual girls, 3 individual boys, and 1 pair of boys.

So for each of those 12 ways that we can put the 3 girls and the 2 boys
together, we can seat the remaining 8 THINGS around the girls in 8! ways.

That's (3!)(2!)(8!) = (6)(2)(40320) = 483840 ways to put the 3 girls together
and the 2 boys together.

3. Now we subtract the result of 2 from the result of 1.

2177280 - 483840 = 1693440 ways.

Edwin