From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?
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Answer Posted / Ankit Kumar Srivastava
To find the number of combinations, we need to calculate the total combinations without restrictions and subtract the combinations that do not have either a green or a blue ball. Let's denote the number of combinations as G (green), B (blue), R (red). We know that G + B + R = 12 (total balls). Also, we have G ≥ 1 and B ≥ 1. Since there are 5 green balls and 4 blue balls, a minimum combination would be 1 green ball, 1 blue ball, and the remaining 6 balls can be any color (R = 6 - 3 = 3 red balls). This gives us 12 - (3 + 1) = 8 combinations for the maximum case with only one of each color. Using the combination formula C(n, k) = n! / [k!(n-k)!], we can calculate all possible combinations without restrictions and subtract the minimum case: Total Combinations = C(12, 0) + C(12, 1) + ... + C(12, 12), Minimum Case Combinations = C(12-3-1, 6) + C(12-4-1, 5) + C(12-5-1, 4). The answer is Total Combinations - Minimum Case Combinations.
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