9 handshakes if everyone shakes everyones hand once
if there are 2 people in a room and each one shakes hands once with every other person in the room, how many hand shakes are there?... answer( 1 handshake) pretty easy isn't it? if there are 3 people in a room and everyone shakes hands with everyone else, how many hand shakes are there? answer( three handshakes) now how many handshakes will there be for 5 people in a room? its your time to shyne...
4*3/2 = 6 handshakes.
9 handshakes correct answer is 10
Depends what you mean, if you mean if everyone shakes hands just once then N-1 handshakes are made. If you mean if everyone shakes hands with everyone else then the answer is (N-1)+(N-2)+....+2+1 (we dont include N as they're not going to shake their own hand, obviously) written as Σn-1i=1 i, this is a arithmetic progression and so the total number of handshakes will be equal to (1+(n-1))(n-1)/2
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.
If that happens you have to times ninexten and the answer would be 90 handshakes
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.
If it doesn't matter who starts it, then there are 60 separate hand-shakes.If it makes a difference who starts the greeting, then there are 120 different cases.
Type your answer here... 6
If there are seven people, then the number of handshakes is 7*6/2 = 21
there would be 26 handshakes if they were all done at once
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.
If there are 6 people in a room, and each person shakes hands with every otherperson in the room, then there will be 15 separate and distinct handshakesbetween different pairs of people.
Person A shakes with nine people (B through J).Person B shakes with eight (C through J because he's already shaken hands with A). Person C shakes with seven, D with six, all the way down to person J, who has already shaken with everyone else. So, 9+8+7+6+5+4+3+2+1+0=45 shakes.
If each person shakes the hand of every other person just once, then 50*49/2 = 1225
The answer is 21 handshakes because the first person shakes hands with the other 6 people. The second person shakes hands with 5 people because they already shook hands with the first person. The third person shakes hands with 4 people because they already shook hands with the first and second person. The fourth person shakes hands with 3 people because they already shook hands with the first, second, and third. The fifth person shakes hands with 2 people because they already shook hands with the first, second, third, and fourth person. The sixth person shakes hands with the seventh person because the rest have already shaken hands with them. The seventh person doesn't have anyone else to shake hands with. Therefore the answer is 21 handshakes.
There are 28 handshakes if everyone shakes hands with everyone else. It is an arithmetic progression and can be solved with the equation Sn=(1+(n-1))(n-1)/2 where Sn is the total sum of handshakes for n people. NB: I have used n-1 instead of n in the equation for the sum of an arithmetic progression, because you're not really going to shake hands with yourself, so you don't include the nth term, in this case 8. Alternatively, you can solve this geometrically by drawing an octagon, drawing lines between all the vertices, and then adding all the lines up.
Assuming that each person shakes hands with every other person, there are 12 people. Let n be the number of people. Then each person shakes hands with (n-1) people and if you ask every person how many hand shakes they made and total them you will get a total of n(n-1) handshakes. However, each handshake involves two people and has been counted twice - once by each person that shook hands - thus number of hand shakes is half of this, giving: n(n-1)/2 = 66 ⇒ n(n-1) = 132 ⇒ n2 - n - 132 = 0 ⇒ (n - 12)(n + 11) = 0 ⇒ n = 12 or -11 You can't have -11 people, therefore there are 12 people.
the first shakes 8 people's hands (remember, not his own), the second 7 (he doesn't shake the first one's hand), then the third shakes six, the fourth shakes 5, the fifth shakes 4, the sixth shakes 3, the seventh shakes 2, and the 8th shakes the 9ths hand so 8+7+6+5+4+3+2+1 = 36