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SobhanR
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There are 11 top managers that need to form a decision group Updated on: 10 Jul 2012, 04:22
Originally posted by SobhanR on 10 Jul 2012, 03:56.
Last edited by Bunuel on 10 Jul 2012, 04:22, edited 1 time in total.
Edited the question and moved to PS subforum
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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
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Bunuel
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There are 11 top managers that need to form a decision group 10 Jul 2012, 04:21
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SobhanR
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
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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.
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There are 11 top managers that need to form a decision group 08 Sep 2018, 08:26
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SobhanR
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
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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
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There are 11 top managers that need to form a decision group 29 Sep 2020, 13:29
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For this question, why can the solution not be found using 2C1* 9C4 [1 out of President or VP attending and 4 from amongst the remaining 9 managers]

Bunuel chetan2u
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There are 11 top managers that need to form a decision group 29 Sep 2020, 19:01
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sahuayush
For this question, why can the solution not be found using 2C1* 9C4 [1 out of President or VP attending and 4 from amongst the remaining 9 managers]

Bunuel chetan2u
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Hi

You are missing out on the teams that do not contain both the president and the Vice President.
So add 9C5 or 126 to it.

Answer = 2C1*9C4+9C5=2*126+126=252+126=378
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There are 11 top managers that need to form a decision group 29 Sep 2020, 19:27
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chetan2u Oh, thank you!

Posted from my mobile device
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There are 11 top managers that need to form a decision group 29 Sep 2020, 20:45
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Generally with these types of Questions that give a Restriction that "2 People can NOT be together" on a team or group, it is better to Find the Total No. of Ways to Choose with NO RESTRICTION ---- and then SUBTRACT out the No. of Ways to Form the Group VIOLATING the RESTRICTION (in which the 2 People are actually together)


For Instance, All the Completely Exhaustive Ways of Making a Team of 5 out of 11 People are the following:



Total No. of Ways to make a Team of 5 out of 11 =

(No. of Ways where P + VP TOGETHER)

+

(No. of Ways P on Team, VP NOT on Team)

+

(No. of Ways VP on team, P NOT on Team)

+

(No. of Ways NEITHER VP nor P on team)



The question says that we want to find the No. of Ways in which P and VP are "NOT TOGETHER" --- this includes the LAST 3 that are ADDED in the List Above.

Rather than Finding all the Different Ways for Each of those LAST 3 Scenarios, it's easier to just SUBTRACT Over the (No. of Ways where P and VP ARE TOGETHER) from the Total Possible Ways when there is NO Restriction.

This will give us the Answer for those LAST 3 Scenarios without having to do as much Calculation.



(Total No. of Ways NO RESTRICTION) - (No. of Ways where P and VP TOGETHER)

= All the Ways in which we can have a team where P and VP are SEPARATED



1st) Total No. of Ways NO RESTRICTION

11 Distinct people in Available Pool
Can Select Any 5

"11 choose 5" = 462 Ways to create a Team of ANY 5 People


-Subtract-


2nd) No. of Ways in which P + VP are TOGETHER (Violating Constraint)

We 1st "Fix" P and VP Together on every different possible Team of 5 --- this has 2 Effects:

-a- We now only have to Select 3 People since every team will have P + VP on it

-b- The Available "Pool" of people from which we can select is reduced by -2 ---- 9 People Remain


"9 Choose 3" = 84 Ways to create a Team with P + VP always Selected Together


462 - 84 = 378


Answer: 378 Ways to choose a Team of 5 where P and VP are SEPERATED
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There are 11 top managers that need to form a decision group 29 Sep 2020, 22:07
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I tried approaching the question like this:
No of ways to fill
the first place: 11, second place: 9 (excluded 1 member, they cannot be on the same team)
3rd 4th and 5th place: 8*7*6
Therefore, the no of ways become: (11*9*8*7*6)/ 5!
5! is the no of repeats that will be present if these people are arranged among themselves.
I did not get an integer value for this.
Bunuel chetan2u could you explain what is wrong in this method?
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