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

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

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- May 26th 2009, 06:23 PMSamprascoinsIn how many different ways can $\displaystyle 8 $ coins be arranged on an $\displaystyle 8 \times 8 $ checkerboard so that no two coins lie in the same row or column?

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

- May 26th 2009, 06:31 PMSampras
- May 26th 2009, 07:43 PMTiRune
- May 26th 2009, 08:18 PMamitface
To summarize, there are 8 possible positions in the first row, 7 in the second, 6 in the third ...

so 8!