The first person must shake hands with 5 other people. The next must shake hands with 4 other people, since you exclude the first person. Keep going and you'll find that there will be 5+4+3+2+1=15 handshakes. Numbers like this are called triangular numbers.
Sixty-six unique, distinct handshakes.
15
Each person will shake hands with every other person, except himself. If there are 25 people, each person will shake hands with 25-1 people, or 24. The number of times each person will shake hands with another, will be 25x24. The number of handshakes will be half of that, as each handshake is between two persons. The formula, in other words, is x(x-1)/2, where x is the number of people. With 25 people, it will be 25x24/2 = 300 handshakes.
Each person shakes hands with every other person at the end of the banquet. When person 1 shakes hands with person 2 that constitutes one handshake even though 2 people are involved. So the answer is 10 total handshakes because the 1st person will have 4 total handshakes(because he can't shake hands with himself, he has 4 and not 5 total handshakes), and then the 2nd person will have 3 total handshakes (you wouldn't say 4 handshakes because you've already included the handshake between person 1 & person 2 when calculating the first person's number of shakes) and so on for the remaining 3 people. On paper the math would look like this: 4+3+2+1=10 Alternatively: Each person shakes hands with 4 others so the answer looks like 5x4 = 20; However, in Fred shaking with 4 others, he shakes with Charlie, similarly, in Charlie shaking with 4 others he shakes with Fred. Thus the Fred-Charlie handshake has been counted twice (once by Fred, once by Charlie), as have all the handshakes, thus the answer is: 5x4 / 2 = 10.
"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. A very simple formula can be applied here. no. of handshakes= (n(n-1))/2 where n is the no.of people present Another conclusion i think the answer is, is the simplest conclusion you can come up with; at the end of the banquet 10 people shake hands with each other so how many handshakes were passed on? 5 hand shakes were given cause that way 10 people did give a hand shakes and since it takes 2 to give out a handshake 5 hand shakes were given. The phrase "Each other" is inclusive, meaning that every person shakes the hand of every other person at the end of the banquet. And since the handshakes that occur when person 1 shakes person 2's hand and vice versa, are the same handshake, those handshakes only count as one entire handshake. This holds true with every other handshake between every other person at the banquet. With this is mind, there will be 45 handshakes since person 1 will shake 9 other people' hands, then person 2 will shake 8 other people's hands, and so on. It would look like this on paper: 9+8+7+6+5+4+3+2+1=45. Each of the 10 people shakes hands with 9 others. If you multiply that, you are counting each handshake double. Therefore, the calculation is 10 x 9 / 2.
So, there will be 3 handshakes among the 3 people at the party.
Sixty-six unique, distinct handshakes.
15
With 3 people, there are only 3 handshakes: AB, AC, and BC. Where it gets interesting is at a party with, say, 10 people ... 45 handshakes. Or in the US Senate when all 100 Senators are present ... 4,950 handshakes.
3. AB, BC and CA.
There will be 28 handshakes. If you ask each person how many handshakes they had they will tell you 7 making 7 x 8 = 56 handshakes in all. But every hand involves two people, so every handshake has been counted twice, thus there are 56 / 2 = 28 handshakes in all.
15 (15 * 15 - 15)/2 = 105
Everyone shakes hands with 4 other people. Since there are 5 people in the room this would suggest there are 5*4 = 20 handshakes. However, you would then be double counting handshakes: A shaking hands with B and B shaking hands with A is, in reality, only one handshake. Thus there are 5*4/2 = 10 handshakes in all.
190
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Each person will shake hands with every other person, except himself. If there are 25 people, each person will shake hands with 25-1 people, or 24. The number of times each person will shake hands with another, will be 25x24. The number of handshakes will be half of that, as each handshake is between two persons. The formula, in other words, is x(x-1)/2, where x is the number of people. With 25 people, it will be 25x24/2 = 300 handshakes.
If each person shakes the hand of every other person just once, then 50*49/2 = 1225