^nC_r + ^{n-1}C_r + ...... + ^rC_r = | vedclass

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(B) The given expression is a sum of combinations: $S = ^rC_r + ^{r+1}C_r + ^{r+2}C_r + ...... + ^{n-1}C_r + ^nC_r$.We use the Hockey-Stick Identity (

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

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(B) The given expression is a sum of combinations: $S = ^rC_r + ^{r+1}C_r + ^{r+2}C_r + ...... + ^{n-1}C_r + ^nC_r$.We use the Hockey-Stick Identity (also known as the Christmas Stocking Identity),which states that $\sum_{i=r}^{n} {^iC_r} = ^{n+1}C_{r+1}$.Step-by-step application:$1$. Since $^rC_r = 1$ and $^{r+1}C_{r+1} = 1$,we can write $^rC_r$ as $^{r+1}C_{r+1}$.$2$. Using the Pascal's Identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.$3$. Applying this iteratively:$(^{r+1}C_{r+1} + ^{r+1}C_r) + ^{r+2}C_r + ...... + ^nC_r$$= ^{r+2}C_{r+1} + ^{r+2}C_r + ...... + ^nC_r$$= ^{r+3}C_{r+1} + ...... + ^nC_r$$= ^{n+1}C_{r+1}$.Thus,the correct option is $B$.

$^nC_r + ^{n-1}C_r + ...... + ^rC_r$ =

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