Q. Evaluate $^{12} C_{2}$ .

4144 210 Permutations and Combinations Report Error

A

$66$

54%

B

$65$

6%

C

$60$

21%

D

$64$

19%

Solution:

$^{12} C_{2} = \frac{12 !}{( 10 ) ! 2 !} < b r / > \frac{12 \times 11 \times ( 10 ! )}{( 10 ! ) \times 2} = 6 \times 11 < b r / >= 66$