Combinations

[sez780]

In how many different ways can a group of 8 be divided into 4 teams of 2 people each?

[krovvidy]

IMO ... (8c2/4)*(6c2/3)*(4c2/2) = 105

Here's the explanation:-

Choose 2 out of 8 (8c2) ... there can be 4 such teams ... so divide by 4 to avoid dupes ...
Out of remaining 6, choose 2 (6c2) ... there can be 3 such teams ... so divide by 3 to avoid dupes ...
Out of remaining 4, choose 2 (4c2) ... there can be 2 such teams ... so divide by 2 to avoid dupes ... (and after arranging 2 such groups, we've exhausted all the players)

Putting all these together you get (8c2/4)*(6c2/3)*(4c2/2) ...

[NishantG]

I have a query, why do we need to divide by the number of teams left???
i.e 4 in first case then 3

According to me there should not be a need to do this division.

answer should be 8C2*6C2*4C2

Answer should be 2520

[Makumajon]

Hi, it is a set-multisubset problem, with each subset consisting of 2 people.
The answer is 8!/(2!2!2!2!)=2520.

Alternatively,
8C2*6C2*4C2*2C2 = 8!/(2!6!)*6!(2!4!)*4!(2!2!)*2!(2!0!) = 8!/(2!2!2!2!) = 2520.

[krovvidy]

@sez780 ... I am still convinced with my earlier answer & solution i.e., 105 ... what's the Official Answer?

@NishantG & @Makumajon, I think, in your approach you've not eliminated the dupes.

[Makumajon]

krovviddy, in that case, it might be a good idea to divide 4 people into four equal groups and see which approach leads to the right answer.

According to your approach: (4c1/4)*(3c1/3)*(2c1/2)=1
According to our approach: 4!(1!1!1!1!)=24.

Now let's examine. Suppose we have people a, b, c, and d and groups 1, 2, 3, and 4.

1 2 3 4
a b c d
a b d c
a c b d
a c d b
a d b c
a d c b
.........

Now we realize the result must be 24.

[krovvidy]

@Makumajon, in fact, 4 people can be divided into 4 groups, with one in each group, in only ONE way.

See this:-

If we brand the teams, i.e., say there are 4 franchises each expecting to pick one player from the available pool of 4 players; then there are 24 ways:

F1 getting P1; F2 getting P2; F3 getting P3; F4 getting P4
F1 getting P1; F2 getting P2; F3 getting P4; F4 getting P3
.....................
F1 getting P4; F2 getting P3; F3 getting P2; F4 getting P1

But, if we just want to divide 4 players into 4 teams (each with one player) ... they are formed only in one way ... because each player going to a specific team doesn't matter ... i.e., we don't have a specific TARGET where these players go.

In essence, if the question does not specify the TARGET, I suppose, we shouldn't assume one ...

[Makumajon]

But since composition of the groups changes, how can we say that each player going to a specific group does not matter? And how is it possible that all 24 diversified groups is actually one way?

[firstgmat]

i agree with krovvidy that we need to eliminate dupes . also, for 4 people there can be only 4 groups. each group is similar to the other. there are no different groups.

[NishantG]

Hmm..Krovvidy, now I understand the point that you are trying make...
But I still believe that divinding into 4 teams should be dividing into team A,B,C or D.

I think it would be a good idea to have the Official Answer to understand what should be the correct approach..