+91-735-322-1155
FlagBACK
Flag Algebra> Please send the solution to the attached ...
question mark

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

supraja venkatraman , 6 Years ago
Grade 11
anser 1 Answers
Samyak Jain

Last Activity: 6 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
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Other Related Questions on algebra

question mark
What is a binomial expression?
algebra
1 Answer Available

Last Activity: 1 Year ago

question mark
Can we divide two vectors? If this is possible then how?
algebra
1 Answer Available

Last Activity: 1 Year ago

question mark
How do you write “the product of 18 and q” as an algebraic expression?
algebra
1 Answer Available

Last Activity: 1 Year ago

question mark
दहावीच्या आय सी एस इ विद्यार्थ्यांनी पुढे काय करायला पाहिजे
algebra
1 Answer Available

Last Activity: 3 Years ago

question mark
Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely. why HCF and not LCM as “least” keywor dis given?
algebra
1 Answer Available

Last Activity: 3 Years ago