The value of -r-110 r r P r -A- 11-B- 11 -1C- 11 -1D- 11 -11

The correct option is B 11! 1∑_r=1^10r.^rP_r =∑_r=1^10(r+1 1)r! =∑_r=1^10 ( (r+1)! r! )=11! 1!=11! 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

nCr Definitions and Properties

The value of ...

Question

The value of $\sum_{r = 1}^{10} r .^{r} P_{r}$ =

11!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11! - 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

11! + 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11! - 11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

If $t a n^{- 1} \frac{1}{a - 1} = t a n^{- 1} \frac{1}{x} + t a n^{- 1} \frac{1}{a^{2} - x + 1}$ , then x is

10 \sum k = 1 (sin \frac{2 k π}{11} - i cos \frac{2 k π}{11})

if the sum of the zeroes of $9 x^{2} - k x + 2$ is $\frac{11}{9}$ find the value of k

\sqrt{1 \frac{1}{2} - [\begin{matrix} 1 \frac{1}{2} - 1 \frac{1}{2} + (\begin{matrix} 1 \frac{1}{2} - 1 \frac{1}{2} - 1 \frac{1}{4} \end{matrix}) \end{matrix}]}

If the line $\frac{x - 1}{2} = \frac{y - 3}{a} = \frac{z + 1}{3}$ lies in the plane $b x + 2 y + 3 z - 4 = 0$ , then

Join BYJU'S Learning Program