10
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.
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.
9 handshakes correct answer is 10
If there are n people who shake hands with each other exactly once, it can be observed that there are n x (n-1) handshakes. Since each handshake is counted twice here,we divide this by 2. Therefore, total number of handshakes is n(n-1)/2. In the given problem, Given: Total handshakes =66 i.e n(n-1)/2=66 n2-n =132 n2-n-132=0 (n-12)(n+11)=0 n =12 or n= -11 As handshakes cannot be negative we discard 11 . Therefore answer is , 12 people.
Each handshake involved 2 people so 12 guests: Guest A made 11 handshakes, B 10 having already been counted as an A-B handshake, C 9 etc for 2 ppl thr is 1 hand shakefor 3 thr is 3(1+2)for 4 thr is 6(1+2+3)hence12 ppl coz..sum of 1st 11 is 66...
Seven different people get together in a place. Each person should handshake only once with all others. Find the total number of handshakes? the first one handshake with 6 person, and second one handshake with remaining 5, and so on....6+5+4+3+2+1=21
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.
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.
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.
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.
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 there are n people who shake hands with each other exactly once, it can be observed that there are n x (n-1) handshakes. Since each handshake is counted twice here,we divide this by 2. Therefore, total number of handshakes is n(n-1)/2. In the given problem, Given: Total handshakes =66 i.e n(n-1)/2=66 n2-n =132 n2-n-132=0 (n-12)(n+11)=0 n =12 or n= -11 As handshakes cannot be negative we discard 11 . Therefore answer is , 12 people.
9 handshakes correct answer is 10
21 handshakes
Each handshake involved 2 people so 12 guests: Guest A made 11 handshakes, B 10 having already been counted as an A-B handshake, C 9 etc for 2 ppl thr is 1 hand shakefor 3 thr is 3(1+2)for 4 thr is 6(1+2+3)hence12 ppl coz..sum of 1st 11 is 66...
The first person can be selected in one of seven ways. Having selected him, the second can be selected from the remaining six in six ways. So there would appear to be 7*6 ways of selecting the couples shaking hands. But, x shaking y's hand is the same handshake as y shaking x's hand. Thus each handshake is conted twice, So the total number of handshakes is 7*6/2 = 21
Each handshake involves two people. If everyone shook only once then there were 36 x 2 ie 72 guests.