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

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

E

Editor

Sep 18, 2022

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

Difficulty: Easy

A:

^{n-2}C_{2r-1}

B:

^{2n-2}C_{2r-1}

C:

^{n}C_{r}

D:

^{n-2}C_{r}

ID: 63273dbaca83163ac3a12455