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16 Mar 2020, 12:42
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Kritisood
Could someone explain why 15c2 is wrong? which scenario am I missing in this?
Hi Kritisood,
To start, there are 18 total people; using 15c2 would assume that you're taking every "pair" of possible people from 15 total people, which is not what the prompt describes. In addition, since people are NOT supposed to shake hands with other members of the same company that they work for, you can't just choose "any 2" people to shake hands - it has to be 2 people from DIFFERENT companies.
GMAT assassins aren't born, they're made,
Rich
16 Mar 2020, 22:55
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EMPOWERgmatRichC
Kritisood
Could someone explain why 15c2 is wrong? which scenario am I missing in this?
Hi Kritisood,
To start, there are 18 total people; using 15c2 would assume that you're taking every "pair" of possible people from 15 total people, which is not what the prompt describes. In addition, since people are NOT supposed to shake hands with other members of the same company that they work for, you can't just choose "any 2" people to shake hands - it has to be 2 people from DIFFERENT companies.
GMAT assassins aren't born, they're made,
Rich
understood! Thanks a lot for pointing this out.
16 Aug 2020, 12:52
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Hello, everyone. A normal "handshake" problem can be solved using the expression
\(\frac{n(n-1)}{2}\)
in which n is the number of individuals involved. Note that this same expression can be used in "how many unique passes" problems or those involving the number of diagonals if the outer sides are not given.
This problem requires a slight tweaking, since we are told that none of the individuals on each "team" will shake hands with anyone else on their own. Whereas before, we saw n - 1—a person cannot shake his or her own hand—now we need to change the expression to n - 3. Note that this new expression can be used in "how many unique passes" problems in which teammates cannot pass to the people right next to them, as well as questions asking about the number of diagonals if the outer shape is given. Here, if there were four team members per company, we would use n - 4, and so on.
\(\frac{n(n-3)}{2}\)
\(\frac{18(18-3)}{2}\)
\(9(15)=9*10+9*5=90+45=135\)
There will be 135 unique handshakes in this particular scenario. There is nothing wrong with using the combination approach, but this method is a little more streamlined, and since it works every time and has broad applicability, I thought I would share it.
Good luck with your studies.
- Andrew
\(\frac{n(n-1)}{2}\)
in which n is the number of individuals involved. Note that this same expression can be used in "how many unique passes" problems or those involving the number of diagonals if the outer sides are not given.
This problem requires a slight tweaking, since we are told that none of the individuals on each "team" will shake hands with anyone else on their own. Whereas before, we saw n - 1—a person cannot shake his or her own hand—now we need to change the expression to n - 3. Note that this new expression can be used in "how many unique passes" problems in which teammates cannot pass to the people right next to them, as well as questions asking about the number of diagonals if the outer shape is given. Here, if there were four team members per company, we would use n - 4, and so on.
\(\frac{n(n-3)}{2}\)
\(\frac{18(18-3)}{2}\)
\(9(15)=9*10+9*5=90+45=135\)
There will be 135 unique handshakes in this particular scenario. There is nothing wrong with using the combination approach, but this method is a little more streamlined, and since it works every time and has broad applicability, I thought I would share it.
Good luck with your studies.
- Andrew
