Permutations and Combination

Roland2rule wrote:Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons

a. 14
b. 12
c. 11
d. 15
e. 16

OA is B

What is the Mathematical approach to solving this question?

Use the answer choices. Test 'C.' Say there were 11 people in the room. There'd be 10 people left for each of those people to shake hands with. But bear in mind that when Dave shakes hands with Roland, it's not a different scenario than Roland shakes hands with Dave. So we need to divide by 2! to account for the fact that those people are interchangeable once we select them. So 11*10/2! = We want 66. But notice we're close.

Test B. If there are 12 people, there are 11 people left to shake with. And we need to divide by 2! to account for the fact that the people are interchangeable. 12*11/2! = 66. So we're done. B is the answer

Roland2rule wrote:Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons

a. 14
b. 12
c. 11
d. 15
e. 16

OA is B

What is the Mathematical approach to solving this question?

And if you're a fiend for algebra, if we say there are 'x' people in the room, there will be (x-1) people for each to shake hands with. Because the two people engaged in a shake are interchangeable, we need to divide by 2!.

So x*(x-1)/2! = 66
x*(x-1) = 132
x^2 - x - 132 = 0
(x-12)(x+11) = 0
x = 12 or x = -11. We can't have negative people, so x=12. The answer is B