The value of -r-1-10r-rP-r -11-11- - 111- - 111- - 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 ...

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Byju's Answer

Standard XI

Mathematics

nCr Definitions and Properties

The value of ...

Question

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

11!

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11! - 1

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11! + 1

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11! - 11

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Solution

The correct option is B 11! - 1
$\sum_{r = 1}^{10} r .^{r} P_{r} = \sum_{r = 1}^{10} (r + 1 - 1) r!$ $= \sum_{r = 1}^{10} ((r + 1)! - r!) = 11! - 1! = 11! - 1$

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Similar questions

Q. Look at this pattern of numbers and take it forward.

1	=	1 ×	1
121	=	11 ×	11
12321	=	111 ×	111
1234321	=	?

Question 47

Find the missing number.
___ + (-11) +111 = 130

$1 \times 1 = 1 11 \times 11 = 121 111 \times 111 = 12321 1111 \times 1111 = 1234321 w r i t e t h e v a l u e o f 11111 \times 11111 =___________________$

Q. Arrange in descending order: 1, 11, 111, -1, -11, -111, 0

Q. Observe the pattern carefully

11 \times 11 = 121

111 \times 111 = 12321

1111 \times 1111 = ?

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The value of ∑10r=1r.rPr =

The correct option is B 11! - 1∑10r=1r.rPr=∑10r=1(r+1−1)r! =∑10r=1((r+1)!−r!)=11!−1!=11!−1

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

The correct option is B 11! - 1
$\sum_{r = 1}^{10} r .^{r} P_{r} = \sum_{r = 1}^{10} (r + 1 - 1) r!$ $= \sum_{r = 1}^{10} ((r + 1)! - r!) = 11! - 1! = 11! - 1$