If I have $1$ jar of $20$ balls with $10$ red, $5$ blue and $5$ yellow, $1$ jar of $30$ balls with $10$ red, $10$ blue and $10$ yellow and $1$ jar of $50$ balls with $15$ red, $20$ blue and $15$ yellow, but I can only pick three from each jar. How many permutations of balls can I get if the order is important?
As shown in this rough sketch
Attempts: $\frac{20!}{(10!\times5!\times5!)} + \frac{30!}{(10!\times10!\times10!)} + \frac{50!}{(15!\times20!\times15!)} = 7.31\times10^{21}$
Let $|P_1|$ number of permutations of the first jar. Let the permutations be $P_1={(p1,p2,p3)} $ where $p1,p2,p3$ can be red, blue or yellow. The total number of permutations is $|P_1|*|P_2|*|P_3|$ so you should multiply each of your terms not add them.
You miss the probability part of picking one Jar from three. Which is $P(J_1)=P(J_2)=P(J_3)=\frac{1}{3}$. Then $P(J_1\cap \text{permutations of balls})+P(J_2\cap \text{permutations of balls})+P(J_3\cap \text{permutations of balls})$
