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NVAEasyJEE 2023Applications of P&C

JEE Mathematics 2023 Question with Solution

Number of integral solutions to the equation x+y+z=21x + y + z = 21, where x≥1,y≥3,z≥4x \geq 1, y \geq 3, z \geq 4, is _____.

Answer

Correct answer:105

Step-by-step solution

Standard Method

Given: x+y+z=21x + y + z = 21 with x≥1,y≥3,z≥4x \geq 1, y \geq 3, z \geq 4.

Find: The number of integral solutions.

Use variable transformation to convert the constrained variables into non-negative integers:

x′=x−1,y′=y−3,z′=z−4x' = x - 1, \quad y' = y - 3, \quad z' = z - 4

where x′≥0,y′≥0,z′≥0x' \geq 0, y' \geq 0, z' \geq 0.

Then

x′+y′+z′=21−1−3−4=13x' + y' + z' = 21 - 1 - 3 - 4 = 13

Now count the non-negative integral solutions of x′+y′+z′=13x' + y' + z' = 13 using combinations:

(13+3−13−1)=(152)\binom{13 + 3 - 1}{3 - 1} = \binom{15}{2}

Evaluate:

(152)=15×142=105\binom{15}{2} = \frac{15 \times 14}{2} = 105

Therefore, the total number of integral solutions is 105105.

Why the transformation works

Given: Lower bounds on x,y,zx, y, z are not zero.

Find: A counting form suitable for stars and bars.

Subtract the minimum required values from each variable. This counts only the excess above the lower bounds. After removing 1,3,1, 3, and 44 respectively, the remaining sum is

21−(1+3+4)=1321 - (1 + 3 + 4) = 13

So the problem becomes finding the number of non-negative integer triples whose sum is 1313. For three variables, the count is

(13+3−13−1)\binom{13 + 3 - 1}{3 - 1}

which gives

(152)=105\binom{15}{2} = 105

Common mistakes

  • Forgetting to adjust the variables for the lower bounds. Directly applying stars and bars to x+y+z=21x + y + z = 21 is wrong because x,y,zx, y, z are not all non-negative from zero. First set x′=x−1,y′=y−3,z′=z−4x' = x-1, y' = y-3, z' = z-4.

  • Subtracting the constraints incorrectly. The reduced sum is 21−1−3−4=1321 - 1 - 3 - 4 = 13, not 1414 or 1212. Always subtract all three minimum values before counting.

  • Using the wrong combination formula. For the number of non-negative integral solutions of a+b+c=13a + b + c = 13, the correct count is (13+3−13−1)\binom{13+3-1}{3-1}. Do not use (133)\binom{13}{3} or (142)\binom{14}{2}.

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