Pigeonhole problem finding the minimum

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???

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• Your textbook answer is wrong; the correct answer is $49$. What book is that? Feb 20 at 0:31

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.
• The Johnson bound implies that for $35$ people the number of meetings is at most $$\left\lfloor\frac{35}{8}\left\lfloor\frac{34}{7}\right\rfloor\right\rfloor = 17.$$ Feb 26 at 22:10
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 $$\Omega$$.