Choose items from an array, each from a different row and column

Question

Thirty items are arranged in a 6-by-5 array. Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.

The answer key gives 1200, but I get 7200. My thought process is as such: Choose 1 item, which eliminates itself and its column/rowmates from the pool, leaving 20. Choose another, leaving 12. Choose the last item. My formula for this is

$_{30}C{_1}\cdot_{20}C{_1}\cdot_{12}C_1 = 30\cdot20\cdot12 = 7200$.

What am I doing wrong?

Answer 1

In an attempt to avoid a "division by symmetry" argument (or at least disguise such an argument with binomial coefficients) you can do the following:

• Simultaneously choose the three columns used ($\binom{5}{3}$ options)
• Simultaneously choose the three rows used ($\binom{6}{3}$ options)
• Pick which row from those chosen is partnered with the left-most selected column ($3$ options)
• Pick which row from those remaining chosen is partnered with the left-most remaining selected column ($2$ options)
• The final selected row is partnered then with the final remaining selected column ($1$ option)

This gives a final total as $\binom{5}{3}\cdot \binom{6}{3}\cdot 3\cdot 2 = 1200$

Answer 2

You are counting the $3!=6$ orders to choose the three items separately. The question only wants the combinations of three items. I think your approach is a good one, but you need to divide by $3!$ at the endj.