1. ## How many combinations?

Hello all! Hopefully someone can help me with this excercise as I am completely stuck myself, and don't even know where to look for the correct answer....

I have 3 columns, each with 12 slots, containing different numbers of balls.
Column A has 3 balls, giving a total of 220 different combinations of balls within the column itself.
Column B has 4 balls, giving a total of 495 combos within the column.
Column C has 5 balls, giving a total of 792 combos within the column.

I place column 1 and 2 next to each other (parallel). I now have to place the balls in the 2 columns in all possible combinations without placing any balls in direct opposite slots. In other words, if the slots are named A-1, A-2 etc., I can not place a ball in B-1 if A-1 already contains a ball. Remember column A should always contain 3 balls and column B always 4 balls.
How many combos are possible in this scenario, and what is the formula for the calculation?

Next I place column C next to A and B. I still have to find all the possible combinations of balls, now for all the 3 columns without placing any balls in direct opposite slots. If A-1 contains a ball I can not place another ball in neither B-1 nor C-1. Remember column A should always contain 3 balls, column B always 4 balls and column C always 5 balls.
How many combos are possible in this scenario, and what is the formula for the calculation?

2. ## Re: How many combinations?

Originally Posted by novice123
I have 3 columns, each with 12 slots, containing different numbers of balls.
Column A has 3 balls, giving a total of 220 different combinations of balls within the column itself.
Column B has 4 balls, giving a total of 495 combos within the column.
Column C has 5 balls, giving a total of 792 combos within the column.
I place column 1 and 2 next to each other (parallel). I now have to place the balls in the 2 columns in all possible combinations without placing any balls in direct opposite slots. In other words, if the slots are named A-1, A-2 etc., I can not place a ball in B-1 if A-1 already contains a ball. Remember column A should always contain 3 balls and column B always 4 balls. How many combos are possible in this scenario, and what is the formula for the calculation?
Select 7 places from twelve: $\binom{12}{7}$.
The number of ways to rearrange the string $AAABBBB$ is $M=\frac{7!}{3!\cdot 4!}=\binom{7}{3}$.
Now for each selection of rows in part 1, there are $M$ ways to arrange the balls as required.
So we have $\binom{12}{7}\cdot\binom{7}{3}$.