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Permutations and Combinations question

2023 · 1 Feb · Shift 2 · Q44
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Permutations and Combinations question

2023 · 1 Feb · Shift 2 · Q44

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
Number of integral solutions to the equation x+y+z=21x+y+z=21x+y+z=21, where x≥1,y≥3,z≥4x \ge 1,y\ge3,z\ge4x≥1,y≥3,z≥4, is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 105

  1. We need the number of integral solutions of x+y+z=21x+y+z=21x+y+z=21 with constraints x≥1,y≥3,z≥4.x\ge 1,\quad y\ge 3,\quad z\ge 4.x≥1,y≥3,z≥4.

  2. Remove the lower bounds by substitution: x′=x−1,y′=y−3,z′=z−4.x'=x-1,\quad y'=y-3,\quad z'=z-4.x′=x−1,y′=y−3,z′=z−4. Then x′,y′,z′≥0.x',y',z'\ge 0.x′,y′,z′≥0.

  3. Substitute into the equation: x′+1+y′+3+z′+4=21x'+1+y'+3+z'+4=21x′+1+y′+3+z′+4=21 x′+y′+z′+8=21x'+y'+z'+8=21x′+y′+z′+8=21 x′+y′+z′=13.x'+y'+z'=13.x′+y′+z′=13.

  4. Now we count the number of non-negative integral solutions of x′+y′+z′=13.x'+y'+z'=13.x′+y′+z′=13. Using the stars and bars formula, the number of solutions of a+b+c=na+b+c=na+b+c=n with a,b,c≥0a,b,c\ge 0a,b,c≥0 is (n+3−13−1)=(n+22).\binom{n+3-1}{3-1}=\binom{n+2}{2}.(3−1n+3−1​)=(2n+2​). So here, Number of solutions=(13+22)=(152).\text{Number of solutions}=\binom{13+2}{2}=\binom{15}{2}.Number of solutions=(213+2​)=(215​).

  5. Compute: (152)=15⋅142=105.\binom{15}{2}=\frac{15\cdot 14}{2}=105.(215​)=215⋅14​=105.

Therefore, the number of integral solutions is 105.\boxed{105}.105​.

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