Permutations & Combinations
Distributions and Constraints
Grade 11
Question:
<p>Let \(a\) and \(b\) be a variable vector such that \(a\) and \(b\) are positive integers. If \(a + b \leq 12\), then the number of values of \((a,b)\) is</p>
<p>(A) \(\binom{11}{2}\)</p>
<p>(B) \(\binom{14}{2}\)</p>
<p>(C) \(\binom{15}{3}\)</p>
<p>(D) \(\binom{12}{3}\)</p>
Step-by-Step Solution
Key Concept: Use stars and bars method or convert constraint into a counting problem of distributing objects.
<p><strong>Solution:</strong> We need to count positive integer pairs \((a,b)\) with \(a + b \leq 12\)</p><p>This is equivalent to finding non-negative integer solutions to \(a + b + c = 12\) where \(a,b \geq 1\) and \(c \geq 0\)</p><p>Substituting \(a' = a-1, b' = b-1\), we need \(a' + b' + c = 10\) with \(a', b', c \geq 0\)</p><p>Number of solutions = \(\binom{10+3-1}{3-1} = \binom{12}{2}\)</p><p>Actually for \(a + b \leq 12\) with \(a,b \geq 1\): we have \(a \in \{1,2,\ldots,11\}\) and for each \(a\), \(b \in \{1,2,\ldots,12-a\}\)</p><p>Total = \(\sum_{a=1}^{11}(12-a) = 11 + 10 + \cdots + 1 = \binom{12}{2} = \binom{15}{3}\) by stars and bars</p><p>∴ Answer is (C) \(\binom{15}{3}\)</p>
Correct Answer: C