The correct option is B 15
For m=5, ∑5r=0(10i)(205−i)
=(100)(205)+(101)(204)+……+(105)(200)
for m=10, ∑5i=0(10i)(2010−i)
=(100)(2010)+(101)(209)+(102)(208)+……+(1010)(200)
for m=15,∑15i=0(10i)(2015−i)
=(100)(2015)+(101)(2014)+(102)(2013)+……+(1010)(205)
and for m = 20, ∑20i=0(10i)(2020−i)
=(100)(2020)+(101)(2019)+……+(1010)(2010)
Clearly, the sum is maximum for m = 15
Note that 10Cr is maximum for r = 5 and 20Cr is maximum for r = 10. Note that the single term 10C5×20C10 (in case m = 15) is greater than the sum
10C200C10+10C201C9+10C202C8+……
10C208C2+10C209C1+10C2010C0 (in case m = 10).
Also the sum in case m = 10 is same as that in case m = 20