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Math Help - How many combinations?

  1. #1
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    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?

    Thank you in advance for a quick an informative answer!
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  2. #2
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    Re: How many combinations?

    Quote Originally Posted by novice123 View Post
    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?
    I will help you with columns A & B.
    Select 7 places from twelve: \binom{12}{7}.
    Those are the rows that will contain a ball.
    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}.
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