- A$^{n + 1}{C_r}$
- ✓$^{n + 1}{C_{r + 1}}$
- C$^{n + 2}{C_r}$
- D${2^n}$
${ = ^{r + 1}}{C_{r + 1}}{ + ^{r + 1}}{C_r}{ + ^{r + 2}}{C_r} + .....{ + ^{n - 1}}{C_r}{ + ^n}{C_r}$
${ = ^{r + 2}}{C_{r + 1}}{ + ^{r + 2}}{C_r} + .....{ + ^{n - 1}}{C_r}{ + ^n}{C_r}$
${ = ^{r + 3}}{C_{r + 1}} + ......{ + ^{n - 1}}{C_r}{ + ^n}{C_r}$.
On solving similar way, we get
$^{n - 1}{C_{r + 1}}{ + ^n}{C_r}{ + ^n}{C_r}{ = ^n}{C_{r + 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_{r + 1}}$.
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$P:\left|z_2-z_1\right|+\left|z_3-z_2\right|+\ldots+\left|z_{10}-z_9\right|+\left|z_1-z_{10}\right| \leq 2 \pi$
$Q:\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\ldots .+\left|z_{10}^2-z_9^2\right|+\left|z_1^2-z_{10}^2\right| \leq 4 \pi$
Then,
$(A)$ $\hat{j}-\hat{k}$ $(B)$ $-\hat{i}+\hat{j}$ $(C)$ $\hat{i}-\hat{j}$ $(D)$ $-\hat{j}+\hat{k}$
| $Face:$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $Probability:$ | $0.1$ | $0.32$ | $0.21$ | $0.15$ | $0.05$ | $0.17$ |
The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$, is