Q: How do you put 9 checkers in rows of four?

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wait not the discription srry

If I am getting what you are saying, then no. 3 rows of 9 and 2 rows equals 5 rows. 5*9=45 and 7*9=63

The degrees of freedom for any contingency table can be calculated simply by the formula (r-1)x(c-1) where r= the number of rows and c= the number of columns. Thus for a contingency table with four rows and four columns the degrees of freedom are 3x3 = 9.

1 row of 180 - 2 rows of 90 - 3 rows of 60 - 4 rows of 45 - 5 rows of 36 - 6 rows of 30 - 9 rows of 20 - 10 rows of 18 - 12 rows of 15 - 15 rows of 12 - 18 rows of 10 - 20 rows of 9 - 30 rows of 6 - 36 rows of 5 - 45 rows of 4 - 60 rows of 3 - 90 rows of 2 - 180 rows of 1 - total of 18 ways within the limits of the question

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wait not the discription srry

Technically, this is impossible as two checkers will always lie in a row. However, how about like this: ................... . @ ............. ..@ @ ......... ..@ ... @ ..... ..@ @ @ @ . ................... (@ = checker) (. = table top, used to ensure picture stays as designed)

If I am getting what you are saying, then no. 3 rows of 9 and 2 rows equals 5 rows. 5*9=45 and 7*9=63

Yes, with 9 chairs in each row.

The degrees of freedom for any contingency table can be calculated simply by the formula (r-1)x(c-1) where r= the number of rows and c= the number of columns. Thus for a contingency table with four rows and four columns the degrees of freedom are 3x3 = 9.

There are 9 rows of stars in total.

1 row of 180 - 2 rows of 90 - 3 rows of 60 - 4 rows of 45 - 5 rows of 36 - 6 rows of 30 - 9 rows of 20 - 10 rows of 18 - 12 rows of 15 - 15 rows of 12 - 18 rows of 10 - 20 rows of 9 - 30 rows of 6 - 36 rows of 5 - 45 rows of 4 - 60 rows of 3 - 90 rows of 2 - 180 rows of 1 - total of 18 ways within the limits of the question

Either 4 parallel rows of three LED's in series or 3 rows of four series coupled LED's (make sure to connect them correctly (+ to - to + to - ...))

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63

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