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Permutations and Combinations question

2021 · 25 Feb · Shift 1 · Q28
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Permutations and Combinations question

2021 · 25 Feb · Shift 1 · Q28

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The total number of positive integral solutions (x, y, z) such that xyz = 24 is :
  1. A
    36
  2. B
    24
  3. C
    45
  4. D
    30
View written solutionFree

Correct answer: D

  1. We need the number of positive integer ordered triples (x,y,z)(x,y,z)(x,y,z) such that xyz=24.xyz=24.xyz=24.

Since 24=23⋅31,24=2^3\cdot 3^1,24=23⋅31, we distribute the prime powers among x,y,zx,y,zx,y,z.

  1. For the factor 232^323: We need the number of non-negative integer solutions of a1+a2+a3=3,a_1+a_2+a_3=3,a1​+a2​+a3​=3, where aia_iai​ is the power of 222 in x,y,zx,y,zx,y,z respectively.

This count is (3+3−13−1)=(52)=10.\binom{3+3-1}{3-1}=\binom{5}{2}=10.(3−13+3−1​)=(25​)=10.

  1. For the factor 313^131: We need the number of non-negative integer solutions of b1+b2+b3=1.b_1+b_2+b_3=1.b1​+b2​+b3​=1.

This count is (1+3−13−1)=(32)=3.\binom{1+3-1}{3-1}=\binom{3}{2}=3.(3−11+3−1​)=(23​)=3.

  1. These choices are independent, so total number of ordered triples is 10×3=30.10\times 3=30.10×3=30.

  2. Therefore, the total number of positive integral solutions is 30.\boxed{30}.30​.

  3. Checking options:

  • A: 363636 ❌
  • B: 242424 ❌
  • C: 454545 ❌
  • D: 303030 ✅

So the correct option is D.

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