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.
The correct answer is $49$.
Such a configuration of meetings is impossible if there are only $48$ people. Let $n_i$ be the number of meetings attended by the $i^\text{th}$ person; so $\sum_{i=1}^{48}n_i=20\cdot8=160$. Since no two people attend the same two meetings, we must have $$\sum_{i=1}^{48}\binom{n_i}2\le\binom{20}2=190.\tag1$$ Now the quantity $\sum_{i=1}^{48}\binom{n_i}2$ is minimized, subject to the constraint $\sum_{i=1}^{48}n_i=160$, when the natural numbers $n_i$ are as nearly equal as possible, i.e., they are all $3$ or $4$, so we have $$\sum_{i=1}^{48}\binom{n_i}2\ge32\binom32+16\binom42=192\gt190=\binom{20}2$$ contradicting $(1)$.
With $49$ people we can even have $21$ meetings with each meeting attended by $8$ people and no two people meeting together more than once. To see this, consider the projective plane $\mathrm{PG}(2,7)$; it has $57$ points and $57$ lines, $8$ points on each line and $8$ lines through each point, any two lines intersecting in just one point. If we regard the points as "people" and the lines as "meetings", we have $57$ people and $57$ meetings. Now let $\Omega$ be a set of $8$ points such that no three are collinear; such a set exists and is called an oval. By the in-and-out formula, the number of lines incident with at least one point of $\Omega$ is $\binom81\cdot8-\binom82\cdot1=36$. Thus there are $57-36=21$ lines/meetings among the $57-8=49$ points/people not in $\Omega$.
