|
Author |
Message |
|
TAGS:
|
|
|
SVP
Joined: 16 Oct 2003
Posts: 1957
Followers: 2
Kudos [?]:
14
[0], given: 0
|
Here are two question which I was having problem with and AkamaiBrah explained it well. Please try
1. In how many ways can 12 students be partitioned into 3 teams, A, B, and C, so that each team contains 4 students?
2. In how many ways can 12 students be divided into 3 teams of 4 people each?
|
|
|
|
|
|
|
Intern
Joined: 10 Aug 2004
Posts: 40
Followers: 0
Kudos [?]:
1
[0], given: 0
|
12C4+8C4+4C4
495+70+1
566
|
|
|
|
|
|
Senior Manager
Joined: 25 Jul 2004
Posts: 296
Followers: 1
Kudos [?]:
3
[0], given: 0
|
There is a subtly here,
In the first one, the answer is 12C4 * 8C4 * 4C4
How many ways you can pick 4 for team A, times the ways you can pick 4 for team B, times the ways you can pick 4 for team C.
In the second one, the teams are unnamed, hence the following assignment
p1 p2 p3 p4 p5 p6 p7 p8 p8 p10 p11 p12
is the same as
p5 p6 p7 p8 p8 p10 p11 p12 p1 p2 p3 p4
so you need to divide the result of the first question by 6, to account for all the sequences that give you the same result.
|
|
|
|
|
|
Intern
Joined: 10 Aug 2004
Posts: 40
Followers: 0
Kudos [?]:
1
[0], given: 0
|
SigEp....does *=multiplication? Wouldn't you add the combinations together?
|
|
|
|
|
|
Senior Manager
Joined: 25 Jul 2004
Posts: 296
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Definately Multiple. For each assignment to the first group, you multiply by the ways you can form the second (and so on) groups.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
