Relatively simple "combination"

[ellexay]

In a classroom, there are 6 students. We need to divide them into 3 pairs for the purpose of assigning homework. In how many ways can be make such pairs?

• 5
• 6
• 15 (correct answer)
• 24
• 36

Feedback: Assume that the students are named A, B, C, D, E, and F. Let us begin with A. She can be paired up as AB, AC, AD, AE, AF (i.e. 5 pairs). Now let us take B. He can be paired up as BC, BD, BE, BF (i.e. 4 pairs. BA was not counted as that has already been considered as AB in the pairs for A). Likewise, C, D and E will have 3, 2 and 1 new pairs. Hence the total number of pairs that can be made are 5 + 4 + 3 + 2 + 1 = 15. Thus, (C) is the answer.


I am wondering why you can't do the following:
3 (6C2)

Thanks...

[x2suresh]

In a classroom, there are 6 students. We need to divide them into 3 pairs for the purpose of assigning homework. In how many ways can be make such pairs?

• 5
• 6
• 15 (correct answer)
• 24
• 36

Feedback: Assume that the students are named A, B, C, D, E, and F. Let us begin with A. She can be paired up as AB, AC, AD, AE, AF (i.e. 5 pairs). Now let us take B. He can be paired up as BC, BD, BE, BF (i.e. 4 pairs. BA was not counted as that has already been considered as AB in the pairs for A). Likewise, C, D and E will have 3, 2 and 1 new pairs. Hence the total number of pairs that can be made are 5 + 4 + 3 + 2 + 1 = 15. Thus, (C) is the answer.


I am wondering why you can't do the following:
3 (6C2)

Thanks...

=(Select first pair from 6 Group* select second pair from remaining 4 group* select third pair from remaining 2)/3! (order doesn't matter)
6C2*4C2*2C2/3! = 15*6*1/6 =15

[ven4gmat]

Since we have to divide 6 students to 3 pairs and obviously each pair consists of 2 two students, this is all about the number of ways in which we can choose 2 students from 6 students which is 6C2 = 15 and hence there are 15 ways

[ellexay]

Thank you!