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Please send the solution to the attached image as soon as possible Please...

Please send the solution to the attached image as soon as possible
Please... 

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Grade:11

1 Answers

Samyak Jain
333 Points
5 years ago
r . rPr  =   r . r! / (r – r)!  =  r . r! / 0!  = r . r!  
(\because nPr  =  n!/(n – r)! and 0! = 1)
 
Let S = ∑r=110  (r . rPr)   i.e.  S = ∑r=110  (r . r!)
Now, ∑r=1n  (r . r!) = ∑r=1n (r + 1 – 1)r! = ∑r=1n ((r+1)r! – r!)
                             = ∑r=1n ((r + 1)! – r!)  
                             = (n+1)! – 1
using telescoping series.
 
\therefore S = (10 + 1)! – 1   =   11! – 1
       = 11! / (11 – 11)! – 1
i.e. S = 11P11 – 1

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