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Permutations and Combinations question

2012 · Shift 0 · Q41
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Permutations and Combinations question

2012 · Shift 0 · Q41

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:
  1. A
    880
  2. B
    629
  3. C
    630
  4. D
    879
View written solutionFree

Correct answer: D

  1. Since balls of the same colour are identical, a selection is completely determined by how many white, green, and black balls are chosen.

  2. Let

    • www = number of white balls chosen, where 0≤w≤100 \le w \le 100≤w≤10
    • ggg = number of green balls chosen, where 0≤g≤90 \le g \le 90≤g≤9
    • bbb = number of black balls chosen, where 0≤b≤70 \le b \le 70≤b≤7

  3. The number of choices for each colour is:

    • White: 111111 choices (000 to 101010)
    • Green: 101010 choices (000 to 999)
    • Black: 888 choices (000 to 777)

  4. Hence, total number of selections including the empty selection is 11×10×8=880.11 \times 10 \times 8 = 880.11×10×8=880.

  5. But the question asks for selection of one or more balls, so exclude the case where no ball is selected: 880−1=879.880 - 1 = 879.880−1=879.

  6. Therefore, the correct option is 879.\boxed{879}.879​.

  7. Comparing with the stored correct answer:

    • Stored answer: D
    • Option D = 879879879
    • This matches our result.
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