Combinations...Please help!

[Evexous]

Suppose you have a checkers board. (8 squares wide and 8 squares long) and you
have 16 checkers.

1. How many ways can the checkers be placed on the board if there are no restrictions?

The total number of ways the checkers can be placed on the board =

16!
8!8!

=12,870 possible ways

2. Now, suppose you must place 2 checkers in each row. How many ways can this be accomplished?

For each row I know that there can be 2 checkers placed anywhere in the 8 squares. This leaves me 6 empty squares. Therefore, I can say:

8!
(8-6)!

and then I will multiply the outcome with 8, knowing that there are 8 rows.

I am not sure if I am approaching this question correctly.

 

[Dr.Peterson]

In part 1, you calculated the number of ways to choose 8 of 16 items ([MATH]_{16}C_8[/MATH]). Why?

In part 2, you calculated the number of ways to arrange 6 of 8 items for each row([MATH]_8P_6[/MATH]). Why?

Then you multiplied by the number of rows. Why?

 

[pka]

Suppose you have a checkers board. (8 squares wide and 8 squares long) and you have 16 checkers. How many ways can the checkers be placed on the board if there are no restrictions?

There are so many ambiguous aspects in this question. First the checkers are all identical with except they may be red or black.
If they are all red then there \(\mathcal{C}_{16}^{64}=\dbinom{64}{16}\) ways to place them.
If eight are red and rest are black then \(\mathcal{C}_{8}^{64}\cdot\mathcal{C}_{8}^{56}\)