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For $2 \leq r \leq n, (\begin{matrix} n \\ r \end{matrix}) + 2 (\begin{matrix} n \\ r - 1 \end{matrix}) + (\begin{matrix} n \\ r - 2 \end{matrix})$ is equal to:

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$(\begin{matrix} n + 1 \\ r - 1 \end{matrix})$

$2 (\begin{matrix} n + 1 \\ r - 1 \end{matrix})$

$2 (\begin{matrix} n + 2 \\ r \end{matrix})$

$(\begin{matrix} n + 2 \\ r \end{matrix})$

answer is D.

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Detailed Solution

Expression $= C (n, r) + 2. C (n, r - 1) + C (n, r - 2)$

$= [C (n, r) + C (n, r - 1)] + [C (n, r - 1) + C (n, r - 2)]$

$= C (n + 1, r) + C (n + 1, r - 1)$

$= C (n + 2, r)$

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