CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The number of values of r satisfying the equation 69C3r−1−69Cr2=69Cr2−1−69C3r is:

A
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2
Solution:
69C3r169Cr2=69Cr2169C3r
or, 69C3r1+69C3r=69Cr2+69Cr21
or, 70C3r=70Cr2
Here, 3r=r2 or 3r+r2=70 (nCr=nCs then r=s or r+s=n)
or, 3r=r2
or, r(r3)=0
or, r=0 which is not possible.
or r=3
r2+3r70=0
or, (r7)(r+10)=0
or, r=7 or r=10 which is not possible.
The value of r satisfying the equation =3,7
Number of values of r satisfying the equation =2
Hence, B is the correct option.

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