wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of (21C110C1)+(21C210C2)+(21C310C3)+(21C410C4)++(21C1010C10) is:

A
221211
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
221210
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
22029
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
220210
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 220210
(21C1 10C1)+(21C2 10C2)+(21C3 10C3)++(21C10 10C10)
=(21C1+ 21C2++ 21C10)(10C1+ 10C2++ 10C10)

Now, 21C1+ 21C2++ 21C10
=12(21C1+ 21C2++ 21C10+ 21C1+21C2++ 21C10)
=12(21C1+ 21C2++ 21C10+ 21C20+21C19++ 21C11)
=12(2212)=2201

10C1+ 10C2++ 10C10
=2101

Required sum =220210

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Properties of Coefficients
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon