Nitish7 wrote:
SravnaTestPrepWhy are we dividing the 6c2* 4c2 * 2c2 by 3!?
Well,
Just want to add my two cents as to how it has been solved on forum and why has that way been used.
lets mark the students numerically as (1,2,3,4,5,6)
How can we select the groups
Ask you self say if i pick (1,2) will the group be different if i pick (2,1) . No. So order does not matter here .
Now this is how i can pick two students for my first group
(1,2),(1,3),(1,4),(1,5),(1,6), (2,3),(2,4),(2,5),(2,6),(3,4)(3,5)(3,6)(4,5)(4,6)(5,6) So total 15 Ways
Say i pick (1,2)
Now how many are left ( 3,4,5,6). Now how can i pick two students for my second group
(3,4)(3,5)(3,6)(4,5)(4,6)(5,6) So total 6 ways
say i this time pick (3,4)
Now how many are left ( 5,6). Now how can i pick two students for my last group
only one way right (5,6) so i ways
So I pick first group AND second group AND third group
so it will be 15 X 6 X 1 =90 ways .
Why are we dividing by 3! when we use formula:
Does the order of group selection matter . Will it be any different if i selected (3,4) for my first group and then (1,2) for second group and (5,6) the last.
No.
But wait , haven't we considered (3,4)(3,5)(3,6)(4,5)(4,6)(5,6) twice . Once in first group and once in second group. So there are 6 cases which have been considered twice. Now if i had picked (5,6 ) initially i would still have 6 cases which would have been considered twice. [(1,2),(1,3),(1,4), (2,3),(2,4),(3,4)}]To get rid of duplication cases we divide by 6.
( hint in the first step we have already identified the cases in which can form groups)
So we have division by 3! in denominator.
Now lets name the group \(G_1, G_2, G_3\)
In how many ways can i assign the projects
Does the order matter . I mean say the order in which project is assigned is \((P_1,P_2,P_3)\) to (\(G_1,G_2,G_3\)) be any different from saying the order of projects is \((P_1,P_2,P_3)\) to (\(G_2,G_1,G_3\)). Yes the order matters.
SAY Project 1 can be assigned to any of the three groups . So we have 3 ways
Now project 2 can be assigned to any of the remaining two groups after one of the three group has been awarded project 1 . So we have 2 Ways
Project 3 can be assigned only one way
So total 6 ways we can assign project to three groups.
So total number of ways we can do the required is 15 Ways of distributing people in 3 groups X 6 ways of distributing 3 projects among the three formed groups.
= 90
Hope this helps
Probus
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Probus
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