The value of $^{n - 1} C_{r - 2} +^{n - 1} C_{r - 1}$ is

^{n - 2} C_{r}

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^{n - 1} C_{r}

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^{n + 1} C_{r}

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^{n} C_{r - 1}

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Solution

The correct option is D $^{n} C_{r - 1}$
Since, $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Similarly, $^{n - 1} C_{r - 2} +^{n - 1} C_{r - 1} =^{n} C_{r - 1}$

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Similar questions

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

(2 \leq r \leq n),

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

^{n} C_{r} +^{n} C_{r - 1}

For $1 \leq r \leq n$ , the value of $ {}^nC_r +^{{}^{n-1}C_r}+^{{}^{n-2}C_r}.....+...+....+^{{}^rC_r} \space is : $

$^{n} C_{r} +^{n} C_{r - 1}$ is equal to

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