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Permutations and Combinations question

2004 · Shift 0 · Q105
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Permutations and Combinations question

2004 · Shift 0 · Q105

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
  1. A
    8C3{}^8{C_3}8C3​
  2. B
    21
  3. C
    38{3^8}38
  4. D
    5
View written solutionFree

Correct answer: B

  1. Let the number of balls placed in the 3 distinct boxes be x1,x2,x3x_1, x_2, x_3x1​,x2​,x3​.

  2. Since all 8 balls are identical and none of the boxes is empty, we need the number of positive integer solutions of x1+x2+x3=8,x_1+x_2+x_3=8,x1​+x2​+x3​=8, with x1,x2,x3≥1.x_1,x_2,x_3 \ge 1.x1​,x2​,x3​≥1.

  3. Use the stars and bars method.

    The number of positive integer solutions of x1+x2+x3=8x_1+x_2+x_3=8x1​+x2​+x3​=8 is (8−13−1)=(72)=7⋅62=21.\binom{8-1}{3-1} = \binom{7}{2} = \frac{7\cdot 6}{2}=21.(3−18−1​)=(27​)=27⋅6​=21.

  4. Therefore, the number of ways is 21.21.21.

  5. Now check the options:

  • A: 8C3=(83)=56{}^8C_3 = \binom{8}{3}=568C3​=(38​)=56 ❌
  • B: 212121 ✅
  • C: 383^838 counts distributions of 8 distinct balls into 3 boxes, not identical balls ❌
  • D: 555 ❌

Hence, the correct option is B.

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