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(52788)   (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.
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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
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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
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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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