Difference between revisions of "2017 AMC 12B Problems/Problem 25"
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Revision as of 14:52, 7 November 2021
Problem
A set of 
people participate in an online video basketball tournament. Each person may be a member of any number of 
-player teams, but no two teams may have exactly the same members. The site statistics show a curious fact: The average, over all subsets of size 
of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size 
of the set of participants, of the number of complete teams whose members are among those people. How many values , 
, can be the number of participants?
Solution
Solution by Pieater314159
Minor edits by Zeric
Let there be 
teams. For each team, there are 
different subsets of players including that full team, so the total number of team-(group of 9) pairs is
![\[T{n-5\choose 4}.\]](http://latex.artofproblemsolving.com/d/e/a/dea29aebd3e741317d1da6d6d5e0d039c865b66c.png)
Thus, the expected value of the number of full teams in a random set of players is
![\[\frac{T{n-5\choose 4}}{{n\choose 9}}.\]](http://latex.artofproblemsolving.com/4/7/3/473fe9b2ab47679ee71a8d44dd0f224bf8a8d2d3.png)
Similarly, the expected value of the number of full teams in a random set of players is
![\[\frac{T{n-5\choose 3}}{{n\choose 8}}.\]](http://latex.artofproblemsolving.com/7/b/b/7bb2aa4cfe97d1e2b11de995288c1cb51ad54fb1.png)
The condition is thus equivalent to the existence of a positive integer such that
![\[\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.\]](http://latex.artofproblemsolving.com/e/1/1/e11179484370cc6d51c3464865a025e00b72cb72.png)
![\[T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\]](http://latex.artofproblemsolving.com/5/9/f/59fde910a2530687da424bd1f4bfde3bde37b66f.png)
![\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}\]](http://latex.artofproblemsolving.com/d/1/c/d1cfec8977bb6d01c6a6576afb77a0a77fd91f18.png)
![\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}\]](http://latex.artofproblemsolving.com/b/5/3/b53d5744d9e02f1bdad30ba0f1af2b4082f6a644.png)
![\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}\]](http://latex.artofproblemsolving.com/f/a/5/fa5ea4bb436228ff0dd76d5fe38fd75453c0fc46.png)
![\[T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}\]](http://latex.artofproblemsolving.com/a/c/c/acc16b7255557e77c2cb36057a28665397194727.png)
Note that this is always less than 
, so as long as is integral, is a possibility. Thus, we have that this is equivalent to
![\[2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).\]](http://latex.artofproblemsolving.com/8/c/5/8c5c769adf9021769328a6400141bc1d703873a5.png)
It is obvious that divides the RHS, and that 
does iff 
. Also, 
divides it iff 
. One can also bash out that 
divides it in 
out of the 
possible residues 
.
Note that 
so by using all numbers from 
to 
, inclusive, it is clear that each possible residue 
is reached an equal number of times, so the total number of working in that range is 
. However, we must subtract the number of "working" 
, which is 
. Thus, the answer is 
.
Alternatively, it is enough to approximate by finding the floor of 
to get .
Video Solution by Dr. Nal:
https://www.youtube.com/watch?v=2p2qYRWbvV4&feature=emb_logo
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Problem |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
