<p>If \(a, b, c, d\) are odd natural numbers such that \(a + b + c + d = 20\), then find the number of values of the ordered quadruplet \((a, b, c, d)\).</p>
Step-by-Step Solution
Key Concept: Convert the constraint on odd numbers to even numbers using substitution: if a, b, c, d are odd, write them as a=2m+1, b=2n+1, c=2p+1, d=2q+1 where m,n,p,q are non-negative integers. This transforms the problem into finding non-negative integer solutions to a simpler equation.
<p><strong>Step 1:</strong> Express odd natural numbers in terms of non-negative integers.</p><p>Let a = 2m+1, b = 2n+1, c = 2p+1, d = 2q+1, where m, n, p, q ≥ 0</p><p><strong>Step 2:</strong> Substitute into the constraint equation.</p><p>(2m+1) + (2n+1) + (2p+1) + (2q+1) = 20</p><p>2m + 2n + 2p + 2q + 4 = 20</p><p>2(m + n + p + q) = 16</p><p>m + n + p + q = 8</p><p><strong>Step 3:</strong> Count non-negative integer solutions using stars and bars.</p><p>Number of solutions = C(8+4-1, 4-1) = C(11, 3)</p><p><strong>Step 4:</strong> Calculate the binomial coefficient.</p><p>C(11, 3) = (11 × 10 × 9)/(3 × 2 × 1) = 990/6 = 165</p><p>∴ Answer: 165</p>
Correct Answer: 165