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Difficult Math Problem #106 - Combinations

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Difficult Math Problem #106 - Combinations

by 800guy » Wed Mar 07, 2007 10:10 am
If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36

oa coming after some ppl answer with explanations. from diff math doc.
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by ssiva » Wed Mar 07, 2007 1:53 pm
1st person will handshake with 9 other persons
2nd person will handshake with 8 other persons but not the first person
3rs person will handshake with 7 other persons but not the first and second person..

so the total handshakes is 9 + 8 + 7 + 6 .... + 1 = 45

Answer D
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by jayhawk2001 » Wed Mar 07, 2007 5:59 pm
Is the answer 10C2 ? i.e. how many ways can 10 people shake hands
with each other.

10C2 = 45, hence D.
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by gabriel » Thu Mar 08, 2007 2:51 am
for one handshake u need 2 people.... therefore the number of handshakes will be 10c2 = 45....
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oa

by 800guy » Fri Mar 09, 2007 10:49 am
oa:

There are 10C2 ways to pick 2 different people out of 10 people.

10C2 = 10!/2!8! = 45 (D)
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by Badri » Fri Mar 09, 2007 10:59 am
well 10C2 is obviously a correct choice because:

here order does not matter. Person '1' shaking hand with person '2' and person '2' shaking hand with person '1' are same events hence

10P2/(!2) = 10C2
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by Cybermusings » Tue Mar 27, 2007 2:43 am
It's a simple solution:

10 people and each person shakes hand with the other person only once. Each handshake needs 2 people. Hence the combination is 10C2 = 10!/10-2! * 2!

= 10*9/2*1

= 90/2

= 45
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by BTGmoderatorRO » Thu Oct 05, 2017 3:16 pm
for this problem, combination can be applied to solve it. the formula is
nCr = n! / (n-r)! r!
= 10! / (10-2)! 2!
=10! / 8! 2!
=3628800 / 80640
=45 handshakes
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If 10 persons meet at a reunion and each person shakes hands

by GMATinsight » Thu Oct 05, 2017 8:53 pm
800guy wrote:If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36

oa coming after some ppl answer with explanations. from diff math doc.
Method-1:
To cause one match to happen we need two teams and two teams out of 10 can be identified in 10C2 ways hence total number of matches
10C2 = 45

Method -2:
First team plays 9 matches, second team playes 8 matches because it has already played with first, third team plays 7 matches as it has already played matches with first and second team
9+8+7+6+5+4+3+2+1 = 45

Method -3:

Every team gets chances to play 1 match with every other team so every team gets to play 9 matches

Total matches = 10*9

but every team has played with every team twice (once when A plays with B and once when B plays with A etc.)

Total identical matches = 10*9/2 = 45
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by Brent@GMATPrepNow » Mon Oct 30, 2017 7:36 am
800guy wrote:If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36
Another approach:

Once everyone has shaken hands, ask each of the 10 people, "How many people did you shake hands with?"
We'll find that EACH PERSON shook hands with 9 people, which gives us a total of 90 handshakes (since 10 x 9 = 90).

From here we need to recognize that every handshake has been counted TWICE. For example, if Person A and Person B shake hands, then Person A counts it as a handshake, AND Person B also counts it as a handshake. Of course only one handshake occurred.

To account for the duplication, we'll divide 90 by 2 to get 45

Answer: D

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by Scott@TargetTestPrep » Thu Nov 07, 2019 6:39 pm
800guy wrote:If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36

oa coming after some ppl answer with explanations. from diff math doc.
The total number of handshakes is 10C2 = (10 x 9)/2! = 45.

Answer: D

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by [email protected] » Sun Nov 24, 2019 11:28 am
Hi All,

This question can be solved in a couple of ways: a high-concept math approach or a "brute-force" answer that anyone can use. I'll focus on the second method.

Since we have 10 people, who will all shake hands with one another, we know that each pair of people will lead to 1 hand shake (and a person CAN'T shake hands with himself or herself).

If we call the people ABCDE FGHIJ

Person A will shake hands with BCDE FGHIJ = 9 shakes

Person B ALREADY shook hands with A, so they won't shake hands again....
Person B will shake hands with CDE FGHIJ = 8 shakes

Person C ALREADY shook hands with A and B, so they won't shake hands again....
Person C will shake hands with DE FGHIJ = 7 shakes

Notice the pattern 9, 8, 7.....the numbers will shrink by 1 with every letter, so we'll end up with...

9+8+7+6+5+4+3+2+1+0 = 45 total handshakes.

Final Answer: D

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