66 hand shakes on a party. If each person shakes every body's hand, how many people were at that party? (I have the answer as 12 people but through use of elementary logic) Could any one suggest a more mathematical way of handling this?
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elementary logic -> mathematical logic – meonstackexchange Mar 24 '15 at 02:52
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The number of handshakes is n(n-1)/2 where n is the number of people. Each of the n people make n-1 handshakes, so the answer would be n(n-1) but we divide by 2 because we double counted each handshake: we thought that A shaking hands with B is different from the reverse case, but they're not.
Faraz Masroor
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The number of possible handshakes is equal to $\dbinom{n}{2}$. These are the combinations of $n$ over $2$. (With combinations you count the ways of selecting $2$ persons out of $n$ so that order does not matter.) Thus you need to solve the equation $$66=\dbinom{n}{2} \iff 66=\frac{n!}{2!(n-2)!} \iff 66=\frac{n(n-1)(n-2)!}{\phantom{(n-1)}2(n-2)!} \iff 132=n(n-1)$$ which is equivalent to the quadratic equation $$n^2-n-132=0 \iff (n+11)(n-12)=0$$ with roots $n=-11$ (rejected, as $n>0$) and $n=12$ which is accepted.
Jimmy R.
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1And one doesn’t even have to solve the quadratic: $$(n-1)^2<2\binom{n}2<n^2;,$$ so the desired number is $\left\lceil\sqrt{2\cdot66},\right\rceil$. – Brian M. Scott Mar 24 '15 at 00:54
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With n amount of people there are $\frac{1}{2}(n^2 +n)$ amount of handshakes because the amount of handshakes is a triangle number, with base $n$. solving that however you may (quadratic formula, fractionation CTS) gives you solutions $132=n^2+n$, $n^2+n-1380=0$ factorizes to $(n-12)(n+11)$ solution are $+12$ or $-11$. There is no such thing as a negative amount of people so we ignore that solution.
N. F. Taussig
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oskar
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It looks like you meant $n^2 + n - 132 = 0$. Please see this tutorial on how to format mathematics on this site. – N. F. Taussig Mar 24 '15 at 02:04