Multiple choice

There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls, if the number of black balls to be selected is twice the number of yellow balls, is

  1. 6

  2. 7

  3. 8

  4. 9
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B Correct answer
Explanation

Let y, b, g be the number of yellow, black, and green balls. We have y + b + g = 20 and b = 2y. Substituting gives y + 2y + g = 20, so 3y + g = 20. Since g must be at least 0, 3y <= 20, so y can be 0, 1, 2, 3, 4, 5, or 6. Each value of y determines a unique g, resulting in 7 possible combinations.

AI explanation

Let the number of yellow balls selected be x and the number of black balls selected be 2x. Because a total of 20 balls are selected, the number of green balls selected is 20 - 3x. For the number of green balls to remain non-negative, the integer x must satisfy the condition 20 - 3x is greater than or equal to 0. The possible values for x are 0, 1, 2, 3, 4, 5, and 6, which gives exactly 7 valid ways to make the selection.

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