sajib2126 wrote:
Hi Bunnel ,i have not understood the 3C3 part . Why do we need to multiply this 3C3 with 6C3/2! , Can you explain please .
To make it simpler, just imagine you were doing a game of 2v2. 4 dudes - Al, Ben, Chuck, Dan. How many ways can you choose two teams of two? (4c2)*(2c2)=6. You can only get {Al+Ben v Chuck+Dan}, {Al+Chuck v Ben+Dan}, and {Al+Dan v Ben+Chuck}. But wait! Because order does not matter, {Al+Ben v Chuck+Dan} is the same as {Chuck+Dan v Al+Ben} and you actually need to divide the 6 possibilities by the number of groups - 2. Using math, (4c2)*(2c2)/2! = 3.
In this problem, you have 6 dudes - Al, Ben, Chuck, Dan, Evan, Frank. Let's turn this into a game of 3 on 3.
Okay, you need to choose 3 for one of the teams. 6c3=20. So, there are 20 ways to pick 3 dudes from a group of 6. Let's call this first group
Group A.
You have one team now and need to "create" another team. How many ways can you chose 3 dudes from the remaining 3 dudes? 3c3=1 way. Let's call this second group
Group B.Now, think about the 20 ways you could pick 2 groups of 3 dudes from 6. If one team consists of Al, Ben, and Chuck in Group A, this is one of the 20 possible groups for Group A. This means that Group B (the other team) would have to be Dan, Evan, and Frank. [b] However, let's think about the other 19 possibilities. Picking Dan, Evan, and Frank in Group A is one of those other 19 possibilities, and one of the total of 20 possibilities. This would matter if the order mattered, but because you just want two teams - you don't really care which team is Group A and which is Group B -, the order does not matter and these 20 choices actually have duplicates. So, divide 20 by 2 to get 10.
In math speak, (6c3)*(3c3)/2!=10.