There are 20 meetings. each neeting has 8 attendees. However, no pair of attendees appear more than once. What is the minimum number of people? To be more specific once a specific pair of people appear in a meeting, this pair will not appear again in any of the other meetings.
My soluton 8C2=28 So there are at leat 28×20=560 distinct pairs Minimum number of people for there to be 560 distinct pairs= 34 But textbook ans is 35???
As you calculated, the 20 meetings generate 560 pairs of people, and each pair can only appear once in all meetings. Imagine that there was a solution with only 34 people, resulting in 561 possible pairs of people. Then apart from one pair $(p_1,p_2)$, each pair should appear exactly once. This means that there certainly is one person (for example $p_3$) who had meetings with every of the 33 other persons.
However, the meetings of $p_3$ should now partition all 33 people in groups of 7, but clearly this is impossible since 7 doesn't divide 33.
Of course, this reasoning does not imply that a schedule could effectively be made with 35 people.