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Friends want to play tennis

Expert replies
by ashish1354 » Sat Sep 20, 2008 4:53 am
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420
B. 2520
C. 168
D. 90
E. 105


i can come up with the solution 8C2*6C2*4C2*2C2 which is incorrect and solution proposes to divide it by 4 * 3 * 2 * 1 which is what i do not understand. Could someone please explain why do we perform division operation.
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Source: — Problem Solving |

by Ian Stewart » Sat Sep 20, 2008 5:36 am
Think of this question:

A group of eight tennis players will be divided into four teams of two. One team will play in the Olympics, one in Wimbledon, one in the Davis Cup and one in the US Open. In how many different ways can the teams be selected?

Here, the order of the teams themselves clearly matters. If we choose {A,B} to go to the Olympics, and {C,D} to go to Wimbledon, that's clearly different from sending {C,D} to the Olympics and {A,B} to Wimbledon. The answer to this question is exactly the answer you give above:

-you have 8C2 choices for the Olympics team;
-you have 6C2 choices for the Wimbledon team;
-you have 4C2 choices for the Davis Cup team;
-you have 2C2 (one) choice for the US Open team.

Multiply these to get the answer: 8C2*6C2*4C2*2C2.

Note that the question I've just asked above is different from the question in the original post. In this question:

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

the order of the teams does not matter. If we choose, say, these teams:

{A,B}, {C,D}, {E,F}, {G,H}

that's exactly the same set of teams as these:

{C,D}, {A,B}, {G,H}, {E,F}

Because the order of the teams themselves does not matter, we must divide by 4! = 24, the number of different orders we can put the four teams in, because all 24 different orders are in fact the same set of teams.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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In how many ways can the letters of the word 'MISSISIPPI' be arranged?

a) 1260
b) 12000
c) 12600
d) 14800
e) 26800

following is the solution to the problem
Total # of alphabets = 10
so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.
we have 4 Is
3 S
and 2 Ps

Hence # of arrangements = 10!/4!*3!*2!

My question is related to the division rule of combinatorics
in the question above we divided by factorial values of repeat elements (4!*3!*2!)
but in the tennis player problem we divide the total possible combinations by 4!, i need to understand the rule & figure out why we divide by 4! rather than(4!*3!*2!) in case of tennis player problem.

If possible please explain why don't we subtract the repeat groups from total combinations rather than dividing by no of repeat combinations.
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by LalaB » Sun Oct 02, 2011 10:24 pm
@Ian Stewart

I understand your explanation and even fully agree with you :) (in fact, what else I can do!? :))

but I have two questions
-do we have another way of solving this problem? if yes, please clarify.
-when should we use the formula (mn)!/(n!)^m (hope I wrote this formula properly heh)

thnx :)
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