Permutations & Combinations
Pascal's identity
Grade None
Question:
<p>If \({}^nC_r\) denotes the number of combinations of \(n\) things taken \(r\) at a time, then the expression \({}^nC_{r+1} + {}^nC_{r-1} + 2 \times {}^nC_r\) equals</p>
<p>\({}^{n+2}C_r\)</p>
<p>\({}^{n+2}C_{r+1}\)</p>
<p>\({}^{n+1}C_r\)</p>
<p>\({}^{n+1}C_{r+1}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the combination terms using the recurrence relation C(n,r) = C(n-1,r) + C(n-1,r-1), then recognize that the sum telescopes into a binomial coefficient of a higher value.
<p><strong>Step 1:</strong> Apply Pascal's identity: <sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r+1</sub> = <sup>n+1</sup>C<sub>r+1</sub></p><p><strong>Step 2:</strong> Regroup the expression as: <sup>n</sup>C<sub>r+1</sub> + <sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r-1</sub></p><p><strong>Step 3:</strong> Apply Pascal's identity twice:</p><p>(<sup>n</sup>C<sub>r+1</sub> + <sup>n</sup>C<sub>r</sub>) + (<sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r-1</sub>) = <sup>n+1</sup>C<sub>r+1</sub> + <sup>n+1</sup>C<sub>r</sub></p><p><strong>Step 4:</strong> Apply Pascal's identity once more:</p><p><sup>n+1</sup>C<sub>r+1</sub> + <sup>n+1</sup>C<sub>r</sub> = <sup>n+2</sup>C<sub>r+1</sub></p><p>∴ Answer: <sup>n+2</sup>C<sub>r+1</sub> (Option B)</p>
Correct Answer: B