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combination question from the gmatclub lesson

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Senior Manager
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combination question from the gmatclub lesson [#permalink] New post 27 Jun 2006, 11:30
Guys,

this question is from the "combinations lessons" on this website.

http://www.gmatclub.com/content/courses ... ations.php

EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team?

Well, the President and the V President are not supposed to be in the same team. So, we effectively have 10 people to make a group of 5. This gives us

10!/(5!*5!) = 252.

so, 252 teams can be formed by including VP or President. So, the total number of teams = 252*2
= 504


The answer is different than what the author got. Where am i wrong?
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Senior Manager
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 [#permalink] New post 27 Jun 2006, 11:54
You have counted the scenario twice where both President and VPwill not be on the team. So, you need to subtract that from your answer..

#ways in which we can form a group of 5 without considering both VP/P : 9C5 = 126

504 - 126
= 378
This shd match with the original answer.
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 [#permalink] New post 27 Jun 2006, 18:01
i thought about it from a different angle this time.

there are 11 people. there are president, vice president and 9 other people. a team of 5 needs to be formed and president and vice president can't be in the same team.

scenario 1
------------
4 places in the team needs to be filled from 9 people.
1 place needs to be filled from 2 people.

number of ways = 2* (9C4) = 2*126 = 252

scenario 2
------------
5 places in the team to be filled by 9 people. president and v president are booted out ;)

number of ways = 9C5 = 126

total number of ways = 378
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 [#permalink] New post 27 Jun 2006, 18:40
shoonya wrote:
i thought about it from a different angle this time.

there are 11 people. there are president, vice president and 9 other people. a team of 5 needs to be formed and president and vice president can't be in the same team.

scenario 1
------------
4 places in the team needs to be filled from 9 people.
1 place needs to be filled from 2 people.

number of ways = 2* (9C4) = 2*126 = 252

scenario 2
------------
5 places in the team to be filled by 9 people. president and v president are booted out ;)

number of ways = 9C5 = 126

total number of ways = 378


Thats exactly the way I solved it. 2 * 9C4 + 9C5 = 378
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  [#permalink] 27 Jun 2006, 18:40
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