neerajgupta wrote:
AnirudhaS wrote:
neerajgupta wrote:
There are 9 students. Team consists of 3 students.Total number of teams =9/3=3
One teamwill have a couple. So selecting this team means selecting the couple and one more guy. Since the couple is already selected, we need one more guy. This can be done in 7 ways
Now, Six students are remaining. No. of ways of Selecting another team of three = 6C3
Now, 3 students are remaining. No. of ways of Selecting another team of three = 3C3
Total Number of ways = 7x6C3x3C3 =140
Ans. D
Since the order of the remaining teams does not matter we need to divide it by 2!
So final answer would be = \(7 * \frac{6C3 * 3C3 }{ 2!}\)
= 70
Answer:
CThen why not divide by 3! as there three teams and order doesnt matter?
What would happen if there were no couples? Do we divide by 3 then??
To be more clear, the reason for dividing by 2 is because using the combination formula 6!/3!3! results in a double counting.
Let's use a simpler example.
Distribute 4 players into 2 teams of 2. Call the players ABCD.
The combination formula would suggest 4!/2!2! = 6. But, is this accurate ?
AB CD
AC BD
AD BC
The combination formula treats AB and CD for example as 2 distinct teams. But in reality, once AB is selected CD is fixed.
So, need to divide 6 by 2 here to account for this, just as is necessary in the problem.
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