There are 7 people at a party. If each person shakes the hand of everyone exactly once, how many handshakes will take place

Answers

Answer 1

Answer: Call the number of people $n$ .

Looking at the first person, they can shake hands with $n-1$ other people (since they can't shake hands with themselves.)

For the second person, they can shake hands with $n-2$ other people. (They can't shake hands with themselves and they already shook with Person 1, on Person 1's turn.

If we carry on this way we see that the total number of shakes is given by:
$(n-1)+(n-2)+...+2+1$ which simplifies to: $(n(n-1))/(2)$ .

For $n=7$ we then have:
$S=(7(7-1))/(2)=(7(6))/(2)=(42)/(2)=21$
OR
$S=(7-1)+(7-2)+(7-3)+(7-4)+(7-5)+(7-6)=6+5+4+3+3+2+1=21$

Answer 2

Answer: There will be 42 handshakes

Answer 3

Answer: There will be 42 handshakes

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Answers

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