Question 1175241
Vanessa plans to arrange A, B, C, D, E, F, G, and H in a row on the wall. Find
the number of permutations for each of the following.
1. A and B must be placed next to each other.<pre>
Case I:  A is immediately left of B

Then she has only 7 things to arrange:  AB, C, D, E, F, G, H

That's 7! ways.

Case II:  A is immediately right of B

Then she has only 7 things to arrange:  BA, C, D, E, F, G, H

That's also 7! ways.

Answer: 7!+7! = 2*7! = 2*5040 = 10080</pre>

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2. A and B must not be placed next to each other.<pre>

We calculate the number of ways the 8 things can be arranged without
restriction, and then subtract the result of problem 1:

8!-2*7! = 40320-10080 = 30240 

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</pre>3. A, B, and C must be placed next to each other, and at the same time, 
E,F and G must also be placed next to each other.<pre>
A,B, and C can be placed together in 3! or 6 ways.
E,F, and G can be placed together in 3! or 6 ways.

That's 6*6 or 36 ways to group A,B,C and E,F,G

For each of those 36 ways to arrange those two triplets, there are 4 things to
arrange, {A,B,C}, (E,F,G}, D, and H.  That's 4!=24 ways.

Answer 6*6*24 = 864 ways.

Edwin</pre>