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If there are 20 persons in a party and if each two of them shake hands with each other, how many hand shakes happen in the party ?

will it be 20C2 = 20*19 or 19+18+...+1? there are 2 different explanation and both seem to be logically correct i said it 20C2 because suppose there re 4 people. A B C D. so the possible hand shakes are {(AB), (AC), (AD), (BC), (BD), (CD)} =6 = 4c2

another way is 19+18+..+1. can somebody explain? answer is 380 or 190?

Sam
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  • In fact, $\binom {20}{2}$ will also get you the right answer. You just have the wrong formula for it. Correct your formula for combinations and you'll get to 190. – Duncan Jun 10 '14 at 17:30
  • Both answers are correct, and both are equal to 190. See my answer below. (BTW, I see that three answers have appeared and I'm still the only one who's up-voted the question. That gets neglected too often.) ${}\qquad{}$ – Michael Hardy Jun 10 '14 at 17:54

3 Answers3

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First of all there is no Probability here, perhaps Combinatorics/Graph theory is more appropriate.

$6$ is also $\binom{4}{2}$ but more importantly it is $3+2+1$.

Thus in all cases the number of hand shakes is $(n-1)+(n-2)+ \cdots + 1 = n(n-1)/2$. Count person by person, ignoring handshakes with people already counted, to see this.

190 is the answer.

John Fernley
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  • thanks. tag was given automatically. fixed it now. – Sam Jun 10 '14 at 17:29
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    You wrote + \ ... + in a seeming attempt to put a small space between the "+" and the dots. That makes sense given the way things get displayed here, but I changed it to +\cdots+, so it appears as $\text{something}+\cdots+\text{something}$. If you write ... instead of \cdots or \ldots or the like in standard TeX or LaTeX rather than the stripped-down thing called MathJax that is used here, then you'll see $\text{something}+\text{...}+\text{something}$, and be standard punctuation conventions, that is incorrect. So use \cdots and \ldots. ${}\qquad{}$ – Michael Hardy Jun 10 '14 at 17:53
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Every person shakes hands with $19$ persons, so at first sight there are $20\times19=380$ handshakes. But by every handshake two persons are involved. So $380$ is the result of double-counting. There are $190$ handshakes.

drhab
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${}_{20}C_2$ is correct.

$19+18+17+\cdots$ is also correct.

Both are the same number.

In fact for all values of $k$, we have $$ 1+2+3+\cdots+k = {}_{k+1}C_2. $$

Michael Hardy
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