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# There are 11 top managers that need to form a decision group

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Intern
Joined: 03 Jul 2012
Posts: 1
There are 11 top managers that need to form a decision group  [#permalink]

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Updated on: 10 Jul 2012, 03:22
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Hi,

I dont seem to find a way to solve the following problem. Any input from your part in solving the question would be sincerely appreciated. Also what would be the best source to learn Permutation and Combination. I am yet to come across a video or document that has clarity and depth. (Or may be the reason is simply that I am coming back to this topic after 15-18 years.)

The question : 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?

The source doesnt provide answer choices.

Thank You.

SobhanR

Originally posted by SobhanR on 10 Jul 2012, 02:56.
Last edited by Bunuel on 10 Jul 2012, 03:22, edited 1 time in total.
Edited the question and moved to PS subforum
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: There are 11 top managers that need to form a decision group  [#permalink]

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10 Jul 2012, 03:21
SobhanR wrote:
Hi,

I dont seem to find a way to solve the following problem. Any input from your part in solving the question would be sincerely appreciated. Also what would be the best source to learn Permutation and Combination. I am yet to come across a video or document that has clarity and depth. (Or may be the reason is simply that I am coming back to this topic after 15-18 years.)

The question : 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?

The source doesnt provide answer choices.

Thank You.

SobhanR

The number of groups of 5 that DO NOT include P and VP equals to the total number of groups of 5 minus the number of groups of 5 that DO include P and VP: $$C^5_{11}-C^2_2*C^3_9=\frac{11!}{6!*5!}-\frac{9!}{6!*3!}=378$$, where $$C^5_{11}$$ is the total number of ways to chose 5 people out of 11, $$C^2_2=1$$ is the number of ways to choose P and VP, and $$C^3_9$$ is the number of ways to choose the remaining 3 members of the group out of 9 people left.

Check Combinatorics and Probability chapters of Math Book for more on these topic:
math-combinatorics-87345.html
math-probability-87244.html

Combinations questions to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52

Probability questions to practice:
DS: search.php?search_id=tag&tag_id=33
PS: search.php?search_id=tag&tag_id=54

Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html

Hope it helps.
_________________
Director
Joined: 27 May 2012
Posts: 684
There are 11 top managers that need to form a decision group  [#permalink]

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08 Sep 2018, 07:26
SobhanR wrote:
Hi,

I dont seem to find a way to solve the following problem. Any input from your part in solving the question would be sincerely appreciated. Also what would be the best source to learn Permutation and Combination. I am yet to come across a video or document that has clarity and depth. (Or may be the reason is simply that I am coming back to this topic after 15-18 years.)

The question : 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?

The source doesnt provide answer choices.

Thank You.

SobhanR

Total ways to choose 5 out of 11 is 11C5 now president and wise president on the same team , then _ _ 9C3
so 11C5 -9C3=378
_________________

- Stne

There are 11 top managers that need to form a decision group   [#permalink] 08 Sep 2018, 07:26
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