3 urns with 6 balls numbered from 1 to 6. We take 1 ball from each urn. In how many way can we have 3 different numbers?

Question

The sample space is given by $$\Omega =\big\{\{i,j,k\}\mid i\in\{R_1,...,R_6\},j\in \{B_1,...,B_6\}, k\in \{G_1,...,G_6\}\big\},$$ where $R$ mean red, $B$ mean blue and $G$ green (Let say that each color correspond to a urn). And thus $|\Omega |=\binom{6}{1}\binom{6}{1}\binom{6}{1}=6^3$. Now we are interested to the event "each ball has a different number." I know that the answer is $6\times 5\times 4$. But I don't understand what is wrong in my argument :

1 take $\binom{6}{3}$ numbers. Then $\binom{3}{2}$ color. In one color I take $\binom{3}{1}$ ball and in the other one, $\binom{2}{1}$ ball. The last ball is determinated.

So I get $$\binom{6}{3}\binom{3}{2}\binom{3}{1}\binom{2}{1}=6\times 5\times 4\times 3,$$ so I have 3 times to much, and I don't understand where I fail.

Answer 1

Your approach should result in $\binom63\times\binom32\times\binom21$ possibilities.

After choosing the numbers (first factor) you choose $2$ distinct colors out of $3$ to label - let's say - the two smallest numbers. This gives the second factor. Then one of these colors is chosen to be the label of the smallest number. This gives the third factor. After that the labeling is complete.

Answer 2

You're overcounting, because choosing numbers $\{1,2,3\}$ then urns $\{B,G\}$ then colour $1$, then colour $2$ (leaving ball $3$ from the red urn) is the same as choosing $\{1,2,3\}$ then urns $\{G,R\}$ then colour $2$, then colour $3$ (leaving ball $1$ from the blue urn). You count each case $3$ times depending on which urn you leave till last.

Answer 3

If you pick a number from the first urn, you can pick any number so $6$ options.

Afterwards you will pick from the second urn. You can't pick the same number as urn $1$, so $5$ options left.

Then the third urn. You cannot pick the same number as urn $1$ or $2$, so $4$ options left.

So: $6\times5\times4$

Answer 4

I assume balls from all 3 urns are different in some way (e.g. colors). So the sample space is $6^3$. All balls having the same number if obviously 6. What about 2 numbers? To avoid overcounting, we select 2 urns: $\binom{3}{2}$ and we have 6 values for this choice. The number on the ball from the third urn must be different, so 5 values. Hence the total number is $$ 6^3 - 6 - \binom{3}{2} \times 6 \times 5 $$