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Please help in solving problems..

by praveen_gmat » Thu Aug 19, 2010 12:02 am
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?

An engagement team consists of a project manager, team leader, and four consultants. There are 2
candidates for the position of project manager, 3 candidates for the position of team leader, and 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?

Also, please tell if there are any techniques to solve this problem.. I am really stuck with permuations and combinations problem. :-(
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by kvcpk » Thu Aug 19, 2010 12:07 am
praveen_gmat wrote: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?
G,M,P,J,B,C are to be seated in 6 seats.
Total possible ways = 6!
Now, M,J are not to sit together.
So see the cases in which MJ sit together.
assuming MJ as one unit, there are 5units to be placed.
So these can be arranged in 5! ways.
again MJ can be internally arranged as MJ or JM. hence 2 ways.

Hence number of ways in which MJ dont sit together = 6!-2*5! = 4*5! = 480 ways.

Hope this helps!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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by praveen_gmat » Thu Aug 19, 2010 12:09 am
kvcpk wrote:
praveen_gmat wrote: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?
G,M,P,J,B,C are to be seated in 6 seats.
Total possible ways = 6!
Now, M,J are not to sit together.
So see the cases in which MJ sit together.
assuming MJ as one unit, there are 5units to be placed.
So these can be arranged in 5! ways.
again MJ can be internally arranged as MJ or JM. hence 2 ways.

Hence number of ways in which MJ dont sit together = 6!-2*5! = 4*5! = 480 ways.

Hope this helps!!

Thanks .. I understood ..Can you please help how the 2nd problem can be done?

An engagement team consists of a project manager, team leader, and four consultants. There are 2
candidates for the position of project manager, 3 candidates for the position of team leader, and 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?
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by kvcpk » Thu Aug 19, 2010 12:17 am
praveen_gmat wrote: An engagement team consists of a project manager, team leader, and four consultants. There are 2
candidates for the position of project manager, 3 candidates for the position of team leader, and 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?
PM - 2 out of which 1 has to be picked. = 2c1 ways
TL - 3 - out of which 1 has to be picked = 3c1 ways
CS - 7 - out of which 4 have to be picked.
Out of these 4, 2 members should never be on same team.
Let us isolate these 2 members.
Now we have a 5 member team and a 2 member team.
For a team to be formed:
We can pick all 4 members from the 5 set (5c4 ways) OR
we can pick 3 members from the 5 set and 1 member from the 2 set. (5c3 * 2c1 ways)
Either of these two should happen. hence number of ways of choosing consultant =
5c4 + 5c3*2c1

Hence Number of teams that can be formed = 2c1*3c1*(5c4+5c3*2c1)
=6*(5+20) = 150

Hope this helps!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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