If in a room there are 10 persons and if each person has to shake hands with all other then total how many hand shakes will be there

10C2=45

10(10-1)/2

simply formulas: n*(n-1)/2 (n-1)*(n/2)

1. n(n-1)/2 2. 10*9/2 3 90/2 4. 45

should be 90

If you say 90 you are including the repetetions. If first person shakes 9 times, the next one requires only 8 shakes. So it goes like, 9, 8 ,7 ,6...1. Can use the formula n*(n-1)/2/. So we get only 45 shake hands. It is similar to no of squares in the chess board. There squares will apply.

If one person is lefty?

Lock in 90 thanks

Forgot 90 Milk Shakes

"Each other" leaves this very open-ended; that depends on if A shakes with B or A shakes with B & C, OR if A shakes with all the other nine, etc. I would say the answer would have to be one of two: 10 or 100. If each person chooses only one to shake with, it would be ten. IF each person shakes with everyone there, all ten, it would be 100. Since this question is pretty vague, Some people may come to the conclusion that the answer is Either 90 assuming everybody stayed to shake hands with each other meaning the first person shook hands with 9 people and the 2nd person did the same etc etc bringing it to the conclusion that you got 90 handshakes. Another answer towards for people would be 45 being that the first person gave a hand shake to 9 people and then left and then the 2nd person gave a handshake to 8 people n then left etc and etc making it 9+8+7+6+5+4+3+2+1=45.