MCQ
For $2\leq\text{r}\leq\text{n},$ $\Big(\frac{\text{n}+1}{\text{r}}\Big)+\Big(\frac{\text{n}}{\text{r}-1}\Big)+\Big(\frac{\text{n}}{\text{r}-2}\Big)$is equal to-
  • A
    $\Big(\frac{\text{n}+1}{\text{r}}\Big)$
  • B
    $2\Big(\frac{\text{n}+1}{\text{r}-1}\Big)$
  • C
    $2\Big(\frac{\text{n}+2}{\text{r}}\Big)$
  • $\Big(\frac{\text{n}+2}{\text{r}}\Big)$

Answer

Correct option: D.
$\Big(\frac{\text{n}+2}{\text{r}}\Big)$
  1. $\Big(\frac{\text{n}+2}{\text{r}}\Big)$
Solution:
${ }^{n+1} C_r+{ }^n C_{r-1}+{ }^n C_{r-2}$
$={ }^{n+1} C_r+{ }^{n+1} C_{r-1}$
$={ }^{n+2} C_r$

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