Evaluate 6C3
We know that
nCr=n!r!n-r!
Given 6C3
Where n=6,r=3
∴nCr=n!r!n-r!=6!3!6-3!=6!3!3!=6×5×4×3!3!×3×2×1=6×5×43×2×1=1206=20
Hence, 6C3 is 20
Evaluate a3+b3+c3−3abc(a+b+c),where(a+b+c)≠0.