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Permutations and Combinations question

2022 · 28 Jun · Shift 2 · Q24
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Permutations and Combinations question

2022 · 28 Jun · Shift 2 · Q24

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives at least 4 and at most 7 candies, C3 receives at least 2 and at most 6 candies, is equal to :
  1. A
    205
  2. B
    615
  3. C
    510
  4. D
    430
View written solutionFree

Correct answer: D

  1. Set up the equation

Let the number of candies received by C1,C2,C3,C4C_1,C_2,C_3,C_4C1​,C2​,C3​,C4​ be x1,x2,x3,x4x_1,x_2,x_3,x_4x1​,x2​,x3​,x4​ respectively.

Since there are 303030 identical candies, x1+x2+x3+x4=30x_1+x_2+x_3+x_4=30x1​+x2​+x3​+x4​=30 with constraints: 4≤x2≤7,2≤x3≤64\le x_2\le 7, \qquad 2\le x_3\le 64≤x2​≤7,2≤x3​≤6 and x1,x4≥0x_1,x_4\ge 0x1​,x4​≥0.

  1. Count solutions for fixed x2x_2x2​ and x3x_3x3​

For any fixed values of x2x_2x2​ and x3x_3x3​, we have x1+x4=30−x2−x3x_1+x_4=30-x_2-x_3x1​+x4​=30−x2​−x3​ The number of non-negative integer solutions of x1+x4=nx_1+x_4=nx1​+x4​=n is n+1n+1n+1 So for fixed x2,x3x_2,x_3x2​,x3​, the number of distributions is 30−x2−x3+1=31−x2−x330-x_2-x_3+1=31-x_2-x_330−x2​−x3​+1=31−x2​−x3​

  1. Sum over all allowed values of x2x_2x2​ and x3x_3x3​

Thus total number of ways is ∑x2=47∑x3=26(31−x2−x3)\sum_{x_2=4}^{7}\sum_{x_3=2}^{6}(31-x_2-x_3)∑x2​=47​∑x3​=26​(31−x2​−x3​)

There are 444 possible values of x2x_2x2​ and 555 possible values of x3x_3x3​.

So,

=\sum_{x_2=4}^{7}\sum_{x_3=2}^{6}31 -\sum_{x_2=4}^{7}\sum_{x_3=2}^{6}x_2 -\sum_{x_2=4}^{7}\sum_{x_3=2}^{6}x_3$$ Now compute each term: - First term: $$31\cdot 4\cdot 5=620$$ - Second term: $$\left(\sum_{x_2=4}^{7}x_2\right)\cdot 5=(4+5+6+7)\cdot 5=22\cdot 5=110$$ - Third term: $$\left(\sum_{x_3=2}^{6}x_3\right)\cdot 4=(2+3+4+5+6)\cdot 4=20\cdot 4=80$$ Hence, $$620-110-80=430$$ --- 4. **Final answer** The required number of ways is $$\boxed{430}$$ So the correct option is **D**.
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