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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Difficult Math Problem #106 - Combinations
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ssiva
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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
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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jayhawk2001
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Is the answer 10C2 ? i.e. how many ways can 10 people shake hands
with each other.
10C2 = 45, hence D.
with each other.
10C2 = 45, hence D.
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Cybermusings
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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
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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BTGmoderatorRO
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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
nCr = n! / (n-r)! r!
= 10! / (10-2)! 2!
=10! / 8! 2!
=3628800 / 80640
=45 handshakes
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GMATinsight
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Method-1:800guy 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.
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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Another approach: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
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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Scott@TargetTestPrep
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The total number of handshakes is 10C2 = (10 x 9)/2! = 45.800guy 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.
Answer: D
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
