Find the value of each of the following quantities: C(8,4)= C(12,7)= C(12,8)= C(11,1)= C(10,9)= C(8,6)=

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Formula:

$nC_k=\frac{n!}{k!(n-k)!}$

Solution:

C(8,4):

n=8;r=4

$8C_4=\frac{8!}{4!(8-4)!}$ = $8C_4=\frac{8!}{4!(4)!}$ =70

C(12,7): Here n=12, r= 7

$12C_7=\frac{12!}{7!(12-7)!}=792$

C(12, 8) n=12, r= 8

$= \frac{12!}{( 8! (12 - 8)! )}$ =495

C(11,1) n=11, r= 1

$= \frac{11!}{( 1! (11 - 1)! )}$ = 11

C(10,9) : n=10, r= 9

$= \frac{10!}{( 9! (10 - 9)! )}$ = 10

C(8,6): n=8, r= 6

$= \frac{8!}{( 6! (8 - 6)! )}$ = 28