In how many ways can we place 9 identical squares – 3 red, 3 white and 3 blue, in a 3×3 square in a way that each row and each column has squares of each three colours?
Let the colors be $RGB$, let the white=green for simplicity, then the top row has the following cases:
$RGB, RBG, BRG, BGR, GRB, GBR = 6$ cases.
Take the $RGB$ configuration first: we have:
$$R G B $$
Below $R$ there are two possibilities, $G, B$, suppose you choose $G$ so now we have:
$R G B$
$G$ ==> Now there are two choices there: $R, B$ then one choice for the last one, and for the whole row beneath, one choice each.
If you chose $B$ then you would have had:
$R G B$
$B$ ==> Only one choice left $(R)$ and then one again for last, and the one choice for each of the spots in the last row.
In total, there are $2 + 1 = 3$ possible arrangements here, it would have been the same for the other cases ($GBR, RBG$, etc…).
The final answer should be $6(3) = 18$, but the actual answer is $9$?
Best Answer
"In total, there are 2+1=32+1=3 possible arrangements here"
No, you don't add them. You multiply them. 2 choices for the second row and each choice yields 1 choice for the third.
"In total, there are 2x1=2 possible arrangements here"
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There are six ways you can do the first row.
In the second row there are 2 choices of color you can put under the red square in the first row. Call that color x. There is only one choices to put under the x square in the first row (not red and not x). Then there's only one choice for the last square of the 2nd row. So there are 2 choices for the second row.
There is only 1 choice for the third row.
So 12 possibilities.