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

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

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

chetan2u Oh, thank you!

Posted from my mobile device

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

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?