(1) 11C4
nCr =\( \frac{n!}{r!(n−r)!}\)
11C4 \(\frac{11!}{4!(11−4)!}=\frac{11×10×9×8×7!}{4×3×2×1×7! }\)= 330
(2) 9C0
nC0 = 1 परिणाम के अनुसार
∴ 9C0 = 1
वैकल्पिक रीति nCr = \(\frac{n!}{r!(n−r)!}=\frac{9!}{0!(9−0)!}=\frac{9!}{1×9!}\)
∴ = 1
(3) 25C23
nCr = \(\frac{n!}{r!(n−r)!}\)
25C23 \(\frac{25!}{23!(25−23)!}=\frac{25×24×23}{23!×2×1}\)= 300
(4) 8C8
nCn = 1
∴ 8C8
वैकल्पिक रीति 8C8 = \(\frac{8!}{8!(8−8)!}=\frac{8!}{8!×0!}\) = 1