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Bunuel
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neerajgupta
Bunuel
Bunuel
Students in a certain class are about to participate in debate teams, which are composed of 3 students each. How many different combinations are possible, if there are 9 students in the class, and the only married couple in the class must be on the same team?

A. 10
B. 42
C. 70
D. 140
E. 1,680

Are You Up For the Challenge: 700 Level Questions


Bumping for discussion. Destroy this one and get Kudos for a correct solution!!!!

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: C
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neerajgupta
Bunuel
Students in a certain class are about to participate in debate teams, which are composed of 3 students each. How many different combinations are possible, if there are 9 students in the class, and the only married couple in the class must be on the same team?

A. 10
B. 42
C. 70
D. 140
E. 1,680

Are You Up For the Challenge: 700 Level Questions



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: C

Then 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??
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neerajgupta
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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: C

Then 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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Bunuel
Students in a certain class are about to participate in debate teams, which are composed of 3 students each. How many different combinations are possible, if there are 9 students in the class, and the only married couple in the class must be on the same team?

A. 10
B. 42
C. 70
D. 140
E. 1,680

Are You Up For the Challenge: 700 Level Questions
Bunuel Could you please share OE or explain the problem again. I am getting an answer different from OA. Thanks in advance Bunuel

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