There are 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?
Ill solve this using the slot method
Possible conditions with the constraints
7*6*5*4 = 35
4*3*2*1
When constraints are in place-- Lets assume that the 2 consultants are already selected,so what we need to do is to find the possible combinations for the other 2 consultants
5*4= 10
2*1
So 35-10=25 possible combinations where the 2candidates are not on the same team.
My concern is why I cant use the same method in this question below
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many ways different arrangements can the 6 people sit?
So now we have 6 candidates Jan,Marcia,Greg,Peter,Bobby,Cindy
To find arrangements in which Marcia and Jan don't sit next to each other,I first find the number of arrangements in which they sit next to each other so the I can subtract those later from the total possible arrangements.
Lets assume J and M are already seated next to eachother,so for the remaining 4 spots we find the possible arrangements
4*3*2*1=24
Since order matters in which M and J sit next to each other opposed to J and M we found out earlier we multiply by 2 thus making it 48 possible arrangements.Therefore mt answer is 6!-48,but the answer is 6!-240=480 .
Why cant I use this method here??I know what the book says,but I want to know why Im wrong,I find it hard to remember too many things in P&C.
Confusion in Combinatorics methods--Kaplann,Manhattan--HELP!
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- Patrick_GMATFix
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Danny you're almost there.
You found out that there are 48 ways to seat all 6 people once J and M are in two adjacent seats, but forgot to account for the pairs of seats they can be adjacent in.
There are 6 seats, so J&M could be next to each other in seats {1,2}, {2,3}, {3,4}, {4,5} or {5,6}. For each of these possibilities, there are 48 ways for them to seat next to each other. So the total # of arrangements that have them next to each other is 48 * 5, and the answer to the question is 6! - 5*48 = 480.
Cheers,
-Patrick
You found out that there are 48 ways to seat all 6 people once J and M are in two adjacent seats, but forgot to account for the pairs of seats they can be adjacent in.
There are 6 seats, so J&M could be next to each other in seats {1,2}, {2,3}, {3,4}, {4,5} or {5,6}. For each of these possibilities, there are 48 ways for them to seat next to each other. So the total # of arrangements that have them next to each other is 48 * 5, and the answer to the question is 6! - 5*48 = 480.
Cheers,
-Patrick
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So I accounted for just one arrangement where the two are seated in the first 2 seats instead of also considering the 4other arrangements of mj and jm?
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yessir
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