GMAT Club Around The World!!! Three pairs of 2 students each need to

Question

Three pairs of 2 students each need to be formed from a group of 7 students for drafting a manuscript each on Physics, Chemistry, and Math respectively. Overall, in how many different ways can such pairs be made?

A. 5040
B. 840
C. 830
D. 630
E. 360

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Dipu10 · Accepted Answer

from 7 students , a pair of 2 can be formed by 7C2 ways : 21
from remaining 5 students, a pair of 2 can be formed in 5C2 ways : 10
from remaining 3 students, a pair of 2 can be formed in 3C2 ways : 3

total ways : 21*10*3 =630 ways

Bunuel · Answer

Experts' Global Video Official Explanation

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Srivatsan97 · Answer

First 6 needs to be selected from 7 = 7c6=7
Then 2 paris from 6 = 6c2= 15
That 2 pairs can be formed in 6 ways
So total ways= 7×15×6=630

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findingmyself · Answer

From 7 students

1st Pair= 7*6
2nd Pair=5*4 (since after first pair only 5 are remaining)
3rd Pair=3*2 (since after 2 pairs)
Total= 7*6*5*4*3*2= 5040
Now each of this pair can be arranged in 2! ways, three pairs can be arranged in 3*2! ways
Thus divide 5040/(2!*3)= 630

Lucabttgr · Answer

I am pretty sure, that this answere is wrong. I agree that the total way 3 teams can be formed, are 630 ways. But now we also have explicity stated, that those are 3 different teams (3 different subjects). So do I not need to multipy the number of possible Teams, with the Number of possible team Subjects, so *6?

Bunuel · Answer

7C2 * 5C2 * 3C2 already counts ordered triples of pairs (Physics pair, Chemistry pair, Math pair).

In other words, you can read the product as:

7C2 = choose the Physics pair
5C2 = choose the Chemistry pair
3C2 = choose the Math pair

Since the roles (Physics, Chemistry, Math) are already tied to the order of selection, there is no extra 3! needed.

You can easily check this by changing the number of students to 4 (A, B, C, D) and the subjects to two, X and Y:

# 
 Subject X 
 Subject Y 
 
  1 
 AB 
 CD 
 
  2 
 AC 
 BD 
 
  3 
 AD 
 BC 
 
  4 
 BC 
 AD 
 
  5 
 BD 
 AC 
 
  6 
 CD 
 AB

As you can see, 4C2 * 2C2 = 6 already lists all possible ordered assignments (X pair, Y pair). There is no extra factor of 2! to multiply.

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GMAT Club Around The World!!! Three pairs of 2 students each need to

Prep Toolkit

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