Pick Five, Always Sixty-Five

From the grid below, pick any number and cross out its row and its column. From what is left, pick another and again cross out its row and column. Carry on until you have chosen five numbers, one from each row and one from each column. Add them up. $$\begin{array}{ccccc} 1 & 6 & 11 & 16 & 21\\ 2 & 7 & 12 & 17 & 22\\ 3 & 8 & 13 & 18 & 23\\ 4 & 9 & 14 & 19 & 24\\ 5 & 10 & 15 & 20 & 25 \end{array}$$ You always reach $65$, whichever five you took. Why? And how many different selections are there?

Solution

Write the labels $0, 1, 2, 3, 4$ down the left of the grid and $1, 6, 11, 16, 21$ across the top. Then the entry in any cell is exactly the sum of its row label and its column label: row $3$ (label $2$), column $4$ (label $16$) holds $2 + 16 = 18$, and so on throughout.

A valid selection takes one entry from each row and one from each column, so among the five chosen entries every row label appears once and every column label appears once. The total is therefore $$(0 + 1 + 2 + 3 + 4) + (1 + 6 + 11 + 16 + 21) = 10 + 55 = 65,$$ whatever cells were picked.

As for how many selections there are: choosing one column for each row, with no column used twice, is the same as arranging the five columns in order, so there are $5! = 120$ of them. Any grid built as such an addition table of row and column labels has the same charm, and the constant is simply the total of all the labels.