Question:
In an examination, the maximum marks for each of three subjects is \(n\) and that for the fourth subject is \(2n\). The number of ways in which candidates can get \(3n\) marks is
In an examination, the maximum marks for each of three subjects is \(n\) and that for the fourth subject is \(2n\). The number of ways in which candidates can get \(3n\) marks is
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For bounded integer solutions, first count all non-negative solutions using combinations, then apply the inclusion-exclusion principle to remove cases violating the upper limits.
Updated On: Jun 22, 2026
- \(\frac{1}{6}(n+1)^2(5n^2+10n+6)^2\)
- \(\frac{1}{6}(n+1)(5n^2+10n+6)^2\)
- \(\frac{1}{6}(n+1)^2(5n^2+10n+6)\)
- \(\frac{1}{6}(n+1)(5n^2+10n+6)\)
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The Correct Option is D
Solution and Explanation
Step 1: Let the marks in four subjects be defined.
Let the marks obtained in the four subjects be \[ x_1,x_2,x_3,x_4 \] For the first three subjects, \[ 0\leq x_1,x_2,x_3\leq n \] For the fourth subject, \[ 0\leq x_4\leq 2n \] The total marks are \[ x_1+x_2+x_3+x_4=3n \]
Step 2: Count total non-negative solutions without restrictions.
The number of non-negative solutions of \[ x_1+x_2+x_3+x_4=3n \] is \[ {}^{3n+3}C_3 \]
Step 3: Subtract invalid cases for the first three subjects.
For any one of \(x_1,x_2,x_3\), if \[ x_i\geq n+1, \] put \[ x_i'=x_i-(n+1) \] Then the remaining sum becomes \[ 3n-(n+1)=2n-1 \] Number of such solutions is \[ {}^{2n+2}C_3 \] Since this can happen for any of the three subjects, subtract \[ 3{}^{2n+2}C_3 \]
Step 4: Add back double-overlap cases.
If two among \(x_1,x_2,x_3\) exceed \(n\), then the remaining sum is \[ 3n-2(n+1)=n-2 \] Number of such solutions is \[ {}^{n+1}C_3 \] There are \[ {}^3C_2=3 \] ways to choose such two subjects.
So add \[ 3{}^{n+1}C_3 \]
Step 5: Subtract invalid cases for the fourth subject.
For the fourth subject, invalid case is \[ x_4\geq 2n+1 \] Put \[ x_4'=x_4-(2n+1) \] Then the remaining sum becomes \[ 3n-(2n+1)=n-1 \] Number of such solutions is \[ {}^{n+2}C_3 \] So the required number of ways is \[ {}^{3n+3}C_3-3{}^{2n+2}C_3+3{}^{n+1}C_3-{}^{n+2}C_3 \]
Step 6: Simplify the expression.
After simplification, \[ {}^{3n+3}C_3-3{}^{2n+2}C_3+3{}^{n+1}C_3-{}^{n+2}C_3 = \frac{1}{6}(n+1)(5n^2+10n+6) \] Therefore, \[ \boxed{\frac{1}{6}(n+1)(5n^2+10n+6)} \]
Let the marks obtained in the four subjects be \[ x_1,x_2,x_3,x_4 \] For the first three subjects, \[ 0\leq x_1,x_2,x_3\leq n \] For the fourth subject, \[ 0\leq x_4\leq 2n \] The total marks are \[ x_1+x_2+x_3+x_4=3n \]
Step 2: Count total non-negative solutions without restrictions.
The number of non-negative solutions of \[ x_1+x_2+x_3+x_4=3n \] is \[ {}^{3n+3}C_3 \]
Step 3: Subtract invalid cases for the first three subjects.
For any one of \(x_1,x_2,x_3\), if \[ x_i\geq n+1, \] put \[ x_i'=x_i-(n+1) \] Then the remaining sum becomes \[ 3n-(n+1)=2n-1 \] Number of such solutions is \[ {}^{2n+2}C_3 \] Since this can happen for any of the three subjects, subtract \[ 3{}^{2n+2}C_3 \]
Step 4: Add back double-overlap cases.
If two among \(x_1,x_2,x_3\) exceed \(n\), then the remaining sum is \[ 3n-2(n+1)=n-2 \] Number of such solutions is \[ {}^{n+1}C_3 \] There are \[ {}^3C_2=3 \] ways to choose such two subjects.
So add \[ 3{}^{n+1}C_3 \]
Step 5: Subtract invalid cases for the fourth subject.
For the fourth subject, invalid case is \[ x_4\geq 2n+1 \] Put \[ x_4'=x_4-(2n+1) \] Then the remaining sum becomes \[ 3n-(2n+1)=n-1 \] Number of such solutions is \[ {}^{n+2}C_3 \] So the required number of ways is \[ {}^{3n+3}C_3-3{}^{2n+2}C_3+3{}^{n+1}C_3-{}^{n+2}C_3 \]
Step 6: Simplify the expression.
After simplification, \[ {}^{3n+3}C_3-3{}^{2n+2}C_3+3{}^{n+1}C_3-{}^{n+2}C_3 = \frac{1}{6}(n+1)(5n^2+10n+6) \] Therefore, \[ \boxed{\frac{1}{6}(n+1)(5n^2+10n+6)} \]
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