Question

The number of integral solutions of \[ x+y+z=13 \] such that \[ 1\le x\le9,\qquad 0\le y\le9,\qquad 0\le z\le9 \] is

• A. \(105\)
• B. \(66\)
• C. \(69\)
• D. \(83\)

Hint:

For bounded integral solutions, first count unrestricted non-negative solutions and then remove invalid cases using Inclusion-Exclusion.

Answer 1

The Correct Option is C

Concept: Use Stars and Bars along with the Principle of Inclusion and Exclusion.

Step 1: Remove the lower bound on \(x\).
Let \[ x'=x-1 \] Then \[ x'+y+z=12 \] with \[ x',y,z\ge0 \] Total solutions: \[ \binom{14}{2} = 91 \]

Step 2: Subtract solutions violating upper bounds.
Condition \(x'>8\): \[ x''=x'-9 \] \[ x''+y+z=3 \] Solutions \[ \binom{5}{2}=10 \] Condition \(y>9\): \[ y'=y-10 \] \[ x'+y'+z=2 \] Solutions \[ \binom{4}{2}=6 \] Condition \(z>9\): Similarly \[ 6 \]

Step 3: Check intersections.
Any two violations together give negative remaining sum. Hence all intersections contribute zero.

Step 4: Apply inclusion-exclusion.
\[ 91-(10+6+6) = 91-22 = 69 \] \[ \boxed{69} \]