Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If nC12= nC6 then nC2=
Q. If
n
C
12
=
n
C
6
then
n
C
2
=
2182
196
KCET
KCET 2005
Permutations and Combinations
Report Error
A
72
23%
B
153
49%
C
306
18%
D
2556
10%
Solution:
Given that,
n
C
12
=
n
C
6
Or
n
C
n
−
12
=
n
C
6
⇒
n
−
12
=
6
⇒
n
=
18
∴
n
C
2
=
18
C
2
=
2
×
1
18
×
17
=
153