What is $C (n, r) + 2 C (n, r - 1) + C (n, r - 2)$ equal to

C (n + 1, r)

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C (n - 1, r + 1)

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C (n, r + 1)

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C (n + 2, r)

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Solution

The correct option is C $C (n + 2, r)$
$C (n_{1} r) + 2 C (n_{1} r - 1) + C (n, r - 2)$

$=^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$

$=^{n} C_{r} +^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2} . . . . (1)$

Now we know that
$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r} . . . . (2)$

$∴^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$

$^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 1} C_{r - 1}$

$∴ (1)$ becomes
$^{n + 1} C_{r} +^{n + 1} C_{r - 1} =^{n + 2} C_{r}$ [By Property Used in $(2)$ ]

$\Rightarrow C (n + 2, r)$ .

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