Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem C. 1403. (February 2017)

C. 1403. An \(\displaystyle n\)-element set has half as many \(\displaystyle {(k-1)}\) element subsets as \(\displaystyle k\)-element subsets, and \(\displaystyle \frac 74\) times as many \(\displaystyle {(k + 1)}\) element subsets as \(\displaystyle k\)-element subsets. Determine the number of \(\displaystyle k\)-element subsets of the set.

(Proposed by L. Koncz, Budapest)

(5 pont)

Deadline expired on March 10, 2017.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Egy \(\displaystyle n\) elemű halmaz \(\displaystyle k\) elemű részhalmazainak a száma \(\displaystyle \binom nk\).

Így a feladat két állítása alapján a következő két egyenletet írhatjuk fel:

\(\displaystyle 2\cdot\binom{n}{k-1}=\binom nk,\)

\(\displaystyle \frac 74\cdot\binom nk=\binom{n}{k+1}.\)

Kifejtve a képleteket:

\(\displaystyle \frac{2n!}{(k-1)!\cdot(n-k+1)!}=\frac{n!}{k!\cdot(n-k)!},\)

\(\displaystyle \frac74\cdot\frac{n!}{k!\cdot(n-k)!}=\frac{n!}{(k+1)!\cdot(n-k-1)!}.\)

Egyszerűsítve:

\(\displaystyle \frac{2}{n-k+1}=\frac1k,\)

\(\displaystyle \frac74\cdot \frac{1}{n-k}=\frac{1}{k+1}.\)

Átszorozva és rendezve az egyenleteket:

\(\displaystyle 3k-1=n,\)

\(\displaystyle 11k+7=4n.\)

Az \(\displaystyle n\) értékét az első egyenletből beírva a másodikba:

\(\displaystyle 11k+7=12k-4,\)

\(\displaystyle k=11,\)

\(\displaystyle n=3k-1=32.\)

A halmaz \(\displaystyle k\) elemű részhalmazainak a száma \(\displaystyle \binom nk=\binom{32}{11}=129\,024\,480\).


Statistics:

163 students sent a solution.
5 points:118 students.
4 points:22 students.
3 points:7 students.
2 points:3 students.
0 point:13 students.

Problems in Mathematics of KöMaL, February 2017