Maths Olympiad Prep

Track / Stage 5 / 216 of 400 #1296 of 2444

Problem 1296

AIME late
Combinatorics Difficulty 5.5 Prove it HMMT February · United States · 2020

Given an 8×88 \times 8 checkerboard with alternating white and black squares, how many ways are there to choose four black squares and four white squares so that no two of the eight chosen squares are in the same row or column?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Next problem →

Official solution

Solution:

Number both the rows and the columns from 11 to 88, and say that black squares are the ones where the rows and columns have the same parity. We will use, e.g., "even rows" to refer to rows 2,4,6,82, 4, 6, 8. Choosing 88 squares all in different rows and columns is equivalent to matching rows to columns.

For each of the 88 rows, we first decide whether they will be matched with a column of the same parity as itself (resulting in a black square) or with one of a different parity (resulting in a white square). Since we want to choose 44 squares of each color, the 44 rows matched to same-parity columns must contain 22 even rows and 22 odd rows. There are (42)2=62\binom{4}{2}^{2} = 6^{2} ways to choose 22 odd rows and 22 even rows to match with same-parity columns.

After choosing the above, we have fixed which 44 rows should be matched with odd columns (while the other 44 should be matched with even columns). Then there are (4!)2=242(4!)^{2} = 24^{2} ways to assign the columns to the rows, so the answer is (624)2=1442=20736(6 \cdot 24)^{2} = 144^{2} = 20736.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.