Handshakes at a PartyDate: 6/23/96 at 21:27:42 From: Anonymous Subject: Handshakes at a Party If there is a party and every person shakes hands with each other once, and there are 45 handshakes, how many people are there at the party? I don't have a clue how to solve it. Date: 6/24/96 at 5:48:58 From: Doctor Anthony Subject: Re: Handshakes at a Party If there are n people at the party, then each person will shake hands with n-1 other people. So with n people each making (n-1) handshakes, it appears at first sight that there are n(n-1) handshakes. However, each handshake will have been counted twice, i.e. A->B and B->A, so we must divide by 2. Total number of handshakes = n(n-1)/2 Now we are given that there were 45 handshakes in all, so we must solve the equation: n(n-1)/2 = 45 n(n-1) = 90 n^2 - n - 90 = 0 (n-10)(n+9) = 0 From this n = 10 or -9 Clearly the -9 has no meaning in this question, so we conclude that n = 10 Number at the party = 10 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/