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4 April, 16:24

How many ways are there to pick 8 balls from large piles of (identical) red, white, and blue balls plus 1 pink ball, 1 lavender ball, and 1 tan ball?

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  1. 4 April, 17:36
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    Suppose order does not matter, then combinations will consist of 0 to 3 of the singleton colors, and 5 to 8 that are some combination of red, white or blue. This implies that we shall formulate the answer as the sum from k=0 to k=3 of:

    N = (Number of ways to pick k of the single colors) * (number of ways to pick 8-k balls from red, white pr blue)

    The first factor is C (3, k), let's call the second factor A (8-k), where A (n) is the number of combinations or red, white or blue when n balls are selected. Given that we want to express A (n) as a formula in terms of n, let's say there are r red balls in a combination (0≤r≤n). Then there are (n-r) balls that are either white or blue that have been left over. Thus, the number of white balls must be integer from 0 to (n-r), for the universe of (n-r+1) possibilities. Hence, the rest are blue in exactly one way, so A (n) is the sum over r=0 to n:

    A (n) = ∑ (1+n-r)

    A (n) = (n+1) (n+1) - n (n+1) 2

    = (n+1) (2n+2-n) / 2

    = (n+1) (n+2) / 2

    thus

    A (5) = 6*7/2=21

    A (6) = 7*8/2=28

    A (7) = 8*9/2=36

    A (8) = 9*10/2=45

    The total will give:

    N=C (3,0) * A (8) + C (3,1) * A (7) + C (3,2) * A (6) + C (3,3) * A (5)

    N=1*45+3*36+3*28+1*21=258

    The answer is 258
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