Answer :
262 combinations ways are there to pick nine balls from large piles of (identical) red, white, and blue balls plus one pink ball, one lavender ball, and one tan ball
In any order, there are 0 to 3 solid colors and 6 to 9 combinations of red, white, and blue. This means formulating the answer as the sum of k = 0 to k = 3 of:
N = (number of ways to draw k for each color)*(red, white, or blue)
The first element is C(3,k). Let's call the second element A(9 - k). where A(n) is the number of combinations, or red, white, or blue if n balls are selected. We want to express A(n) in terms of n, so let's say there are r red balls (0 ≤ r ≤ n). Then we are left with (n-r) white or blue balls. Therefore, in a universe of (n-r+1) possibilities, the number of white balls must be an integer from 0 to (n-r). So A(n) is the sum from r = 0 to n because the rest are blue in exactly one way.
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(6)=7*7/2 = 24.5
A(7)=8*8/2 = 32
A(8)=9*9/2 = 40.5
A(9)=10*10/2 = 20
The total will give:
N=C(3,0)*A(9)+C(3,1)*A(8)+C(3,2)*A(7)+C(3,3)*A(6)
N=1*20+3*40.5+3*32+1*24.5 =262
The answer is 262.
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