Theres a party with 5 families. One family has 5 members, two have 3 members and two have 2 members. 1. If everyone shakes hands with everyone else, how many handshakes occur?2. If everyone shakes hands with everyone else, except their family members, how many handshakes occur?. Created by ?. education-reference-en - trivia-en

Algebra Trivia Problem HELP!?

J. J..

1. 105 handshakes

2. 87 handshakes

1. number of people = 5+3+3+2+2 = 15

That means everyone shakes 14 hands (not their own so subtract 1 from 15)

14 x 15 = 210 "hands" shaken

Each handshake consists of 2 hands being shaken

210 / 2 = 105 handshakes

Another way to envisage it is to line up all 15 people

person 1 shakes 14 hands and walks away

person 2 then shakes 13 hands (he has already shaken person 1's) and walks away

person 3 then shakes 12 hands (alreadyshaken persons 1 and 2) and walks away

person 4 then shakes 11 hands (already shaken persons 1,2 and 3) and walks away

and so on

14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105 handshakes

The 5 members of family one each shake 10 hands = 5 x 10 = 50 handshakes and walk away

The 3 members of family 2 each shake 3+2+2 = 7 hands and walk away 3 x 7 = 21

The 3 members of family 3 each shake 2+2 = 4 hands and walk away 3 x 4 = 12

The 2 members of family 4 each shake 2 hands and walk away 2 x 2 = 4

The 2 members of family 5 have now already shaken hands with every other family's members

50+21+12+4 = 87 handshakes

Another way of calculating it is

Take the 105 calculated in Part 1

In family 1 AB +AC +AD + AE + BC+ BD + BE + CD + CE + DE = 10 handshakes don't occur

This is 4 + 3 + 2 + 1

In family 2 AB AC BC = 3 handshakes don't occur = 2+1 = 3

In family 3 again 3 handshakes don't occur

In family 4 AB only

In family 5 AB only

10+3+3+1+1 = 18 handshakes just between family members don't occur

105 - 18 = 87 handshakes DO OCCUR

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