Question:
In how many different ways can three persons \(A\), \(B\), \(C\) having \(6\), \(7\) and \(8\) one-rupee coins respectively, donate Rs.\(10\) collectively?
In how many different ways can three persons \(A\), \(B\), \(C\) having \(6\), \(7\) and \(8\) one-rupee coins respectively, donate Rs.\(10\) collectively?
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For restricted integer-solution problems, first count all non-negative solutions using stars and bars, then subtract invalid cases using the inclusion-exclusion principle.
Updated On: Jun 18, 2026
- \(47\)
- \(66\)
- \(56\)
- \(60\)
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The Correct Option is A
Solution and Explanation
Step 1: Form the equation.
Let the contributions of \(A\), \(B\), and \(C\) be \(x\), \(y\), and \(z\) respectively.
Then \[ x+y+z=10 \] with restrictions \[ 0\le x\le 6, \] \[ 0\le y\le 7, \] \[ 0\le z\le 8. \] We need the number of non-negative integer solutions satisfying these conditions.
Step 2: Count all non-negative solutions.
Ignoring the upper bounds, the number of solutions of \[ x+y+z=10 \] is \[ {}^{10+3-1}C_{3-1} = {}^{12}C_2 = 66. \]
Step 3: Subtract solutions violating the restrictions.
First, consider \[ x\ge 7. \] Let \[ x'=x-7. \] Then \[ x'+y+z=3. \] Number of solutions: \[ {}^{3+3-1}C_2 = {}^5C_2 = 10. \] Next, consider \[ y\ge 8. \] Let \[ y'=y-8. \] Then \[ x+y'+z=2. \] Number of solutions: \[ {}^4C_2 = 6. \] Next, consider \[ z\ge 9. \] Let \[ z'=z-9. \] Then \[ x+y+z'=1. \] Number of solutions: \[ {}^3C_2 = 3. \]
Step 4: Check intersections.
For \[ x\ge7,\quad y\ge8, \] we would need \[ x'+y'+z=-5, \] which is impossible.
Similarly, \[ x\ge7,\; z\ge9 \] and \[ y\ge8,\; z\ge9 \] are impossible.
Hence all intersections are zero.
Step 5: Apply inclusion-exclusion.
Therefore, \[ 66-(10+6+3) = 66-19 = 47. \]
Step 6: Final conclusion.
Hence, the required number of ways is \[ \boxed{47}. \]
Let the contributions of \(A\), \(B\), and \(C\) be \(x\), \(y\), and \(z\) respectively.
Then \[ x+y+z=10 \] with restrictions \[ 0\le x\le 6, \] \[ 0\le y\le 7, \] \[ 0\le z\le 8. \] We need the number of non-negative integer solutions satisfying these conditions.
Step 2: Count all non-negative solutions.
Ignoring the upper bounds, the number of solutions of \[ x+y+z=10 \] is \[ {}^{10+3-1}C_{3-1} = {}^{12}C_2 = 66. \]
Step 3: Subtract solutions violating the restrictions.
First, consider \[ x\ge 7. \] Let \[ x'=x-7. \] Then \[ x'+y+z=3. \] Number of solutions: \[ {}^{3+3-1}C_2 = {}^5C_2 = 10. \] Next, consider \[ y\ge 8. \] Let \[ y'=y-8. \] Then \[ x+y'+z=2. \] Number of solutions: \[ {}^4C_2 = 6. \] Next, consider \[ z\ge 9. \] Let \[ z'=z-9. \] Then \[ x+y+z'=1. \] Number of solutions: \[ {}^3C_2 = 3. \]
Step 4: Check intersections.
For \[ x\ge7,\quad y\ge8, \] we would need \[ x'+y'+z=-5, \] which is impossible.
Similarly, \[ x\ge7,\; z\ge9 \] and \[ y\ge8,\; z\ge9 \] are impossible.
Hence all intersections are zero.
Step 5: Apply inclusion-exclusion.
Therefore, \[ 66-(10+6+3) = 66-19 = 47. \]
Step 6: Final conclusion.
Hence, the required number of ways is \[ \boxed{47}. \]
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