## A Counting Puzzle Raffle – Solution

Many thanks to all who participated in our counting puzzle raffle.

The winner was selected randomly from correct entries and has requested to remain anonymous.

Here’s a solution.

Many people used the hook-length formula to compute the answer. It might be interesting to note that all the people who did not use the hook-length formula got the correct answer, but only half of those who did use the formula were correct.

Since you can read about the hook-length formula elsewhere, here we’ll deduce the answer using basic counting techniques.

Instead of using word titles, we’ll use the numbers 1 through 10 and put them into the diagram so that the numbers increase left to right across any row and down any column.

The number 1 must go in the upper left square.

Now, let’s fill in squares until the square to the right of the 1 is filled, then stop.  There are many ways to proceed, so we split into cases. The diagram below shows all the cases. The first number that is put into the second column must go at the top of that column, which is the square to the right of the 1, and that’s when we stop. So we end up with 4 cases. In each of the 4 cases, notice that we are free to choose any 3 numbers from the numbers that haven’t been placed yet for the remaining 3 squares in row 1. These numbers must be placed in row 1 in ascending order, and the numbers still remaining must be placed in the unfilled squares in the first 2 columns. The number of ways the unfilled squares in the first 2 columns can be filled is quite limited and can be counted directly.

Case 1: $\binom{8}{3} \cdot 5 = 280$.

Case 2: $\binom{7}{3} \cdot 5 = 175$.

Case 3: $\binom{6}{3} \cdot 3 = 60$.

Case 4: $\binom{5}{3} \cdot 1 = 10$.

Adding these up, we get 280 + 175 + 60 + 10 =  525, which is the answer. 