# coins

• May 26th 2009, 07:23 PM
Sampras
coins
In how many different ways can $8$ coins be arranged on an $8 \times 8$ checkerboard so that no two coins lie in the same row or column?

So it would be $64 \times 57 \times 52 \times 49$?
• May 26th 2009, 07:31 PM
Sampras
Quote:

Originally Posted by Sampras
In how many different ways can $8$ coins be arranged on an $8 \times 8$ checkerboard so that no two coins lie in the same row or column?

So it would be $64 \times 57 \times 52 \times 49$?

This would be for a $4 \times 4$ board. But is the reasoning correct?
• May 26th 2009, 08:43 PM
TiRune
• May 26th 2009, 09:18 PM
amitface
To summarize, there are 8 possible positions in the first row, 7 in the second, 6 in the third ...

so 8!