SOLUTION: 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.

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

Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

I solved part (1) yesterday under this link

https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1175182.html

https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1175182.html



But I agree to reproduce it here  AGAIN,  because it is closely connected with part  (2).


                    Part (1)


For the rest of  6  letters,  C, D, E, F, G and H,  there are   6! = 720 possible permutations.

For each of these  720  permutations,  we have  7  positions to place the blocked  AB  or  BA.

THEREFORE,  the full number of possible arrangements,  satisfying the posed conditions, is

            6!*2*7 = 2*7! = 10080.             ANSWER

Solved.


                    Part  (2)


These permutations are  COMPLEMENT to   8! = 40320,

so their number is   40320 - 10080 = 30240.           ANSWER

Solved.

-------------


I will not solve here part  (3),  since I think it is  TOOOOOO  much for one post.

It is not a way to teach people.


It is the way to work instead of them and to perform their job,  what I do not want to do.



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
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.
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

-----------------------------------------
2. A and B must not be placed next to each other.

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 

------------------------------------------------------------
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.
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

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