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Students in a certain class are about to participate in debate teams

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Students in a certain class are about to participate in debate teams  [#permalink]

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New post 07 Jul 2020, 11:25
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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
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Students in a certain class are about to participate in debate teams  [#permalink]

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New post 07 Jul 2020, 14:32
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams \(= \frac{7C1*6C3*3C3}{2! }= 70 \) [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn wrote:
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
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 07 Jul 2020, 18:06
Could you please elaborate why have you added 3C3 in this combination set?

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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 07 Jul 2020, 18:20
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After selecting the team that includes married couple, we left with 6 students.

Number of ways to select second team = Number of ways to select 3 students out of 6= 6C3

Now we left with 3 students.

Number of ways to select third team= 3C3 = 1





Shresta5 wrote:
Could you please elaborate why have you added 3C3 in this combination set?

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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 11 Jul 2020, 02:25
tamarlvn wrote:
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


Make two cases:
Case 1 - Team that have married couple
Case 2 - Teams without married couple

Case 1 * Case 2 -> Our three desired team

Case 1 -> Ways to select first two member of team -> Only one way (Married couple) -> Ways to select third team member out of 7 remaining member -> 7
Hence, number of ways to select team 1 -> 1*7 = 7

Case 2 ->
Number of ways to select second team or T2 of three member -> 6c3 = 20
Number of ways to select third team or T3 of three member -> 3c3 = 1

Since the order of selection does not matter in team selection, our cases will be reduced by half
Because first combination -> [ T2-(m1,m2,m3) T3-(m4,m5,m6)] is same as second combination -> [ T2-(m4,m5,m6) T3-(m1,m2,m3)]
Hence, possible combinations will be 20/2=10

Thus, total number of cases will be = Case 1 * Case 2
= 7*10
= 70
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 14 Jul 2020, 19:08
Bunuel need your answer
I'm getting 140.

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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 14 Jul 2020, 23:17
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams \(= \frac{7C1*6C3*3C3}{2! }= 70 \) [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn wrote:
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



Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

Thanks in advance.
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 15 Jul 2020, 04:28
Sir nhi bhai chalega!

Since we can't differentiate between 2 groups in which we don't have married couple, we have to divide by 2!. Logic is pretty similar to the following question.

Find the number of arrangements of the letters of the word "OFF"?

rishit924 wrote:
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams \(= \frac{7C1*6C3*3C3}{2! }= 70 \) [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn wrote:
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



Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

Thanks in advance.
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 15 Jul 2020, 05:45
Shresta5 wrote:
Could you please elaborate why have you added 3C3 in this combination set?

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Think it in the way that you have now only three members left to be picked and you have to make the third team of three people.

Let's say you have three person A, B and C left and you have to make a team of three people comprising of A, B and C.

Now think of possible ways you can select the team from these people ->
Way 1 -> You select A first, then B and finally C -> Final result is ABC
Way 2 -> You select B first, then C and finally A -> Final result is BCA
Way 3 -> You select C first, then A and finally B -> Final result is CAB
Way 4 -> You select A first, then C and finally B -> Final result is ACB
Way 5 -> You select B first, then A and finally C -> Final result is BAC
Way 6 -> You select C first, then B and finally A -> Final result is CBA

But if you'll notice, you are getting only one actual combination and order does not matter as team consist of only those three people.

Thus, there is only one way to select such team and it can be represented as 3C3 -> Selection of 3 from a set of 3
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 15 Jul 2020, 21:30
1
The OA is 70. Option (C). The major confusion is between 140 and 70. I'll try explaining.

For the married couple to remain together, consider them as one entity. There are 7 other people in the class.
If anyone member gets along with the couple, those 3 can form a debate team.

So the number of ways in which a team can be formed containing the married couple = 7C1 = 7.

Now the remaining 6 people need to be arranged in groups of 3.
Therefore, selecting 3 people out of 6 = 6C3 = 20 ---- (A)

Let us consider a case. Members are named A - G and the couple are simply H and W.

Example case :

H, W, A are together --- Team 1 ( 7C1)
B, C, D are together --- Team 2 (6C3)
E, F, G are together --- Team 3

Now when B, C, and D are selected in 6C3; E, F, and G are automatically selected as well for Team 3. There would also be one other case where you'll select E, F, and G for Team 1. In that case, B, C, and D would form Team 3.

Carefully understand that both the cases are identical since the order of teams doesn't matter. That's why we divide by 2!.

Hence, (6C3*7C1)/2! = 70.

Option (C)
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Re: Students in a certain class are about to participate in debate teams  [#permalink]

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New post 15 Jul 2020, 21:33
rishit924 wrote:
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams \(= \frac{7C1*6C3*3C3}{2! }= 70 \) [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn wrote:
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



Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

Thanks in advance.



I hope you find the above explanation helpful!
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Re: Students in a certain class are about to participate in debate teams   [#permalink] 15 Jul 2020, 21:33

Students in a certain class are about to participate in debate teams

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