Question

Evaluate:
$\frac{n!}{r!(n-r)!}$
for (i) n = 10, r = 4
(ii) n = 15, r = 15

Answer 1

Hello dear,

● Answer-
(i) 210
(ii) 0

[Refer to image for explaination]

● Permutation -
It is defined as arrangement of r things that can be done out of total n things. This is denoted by nPr which is equal to n!/(n-r)!.

● Combination -
It is defined as selection of r things that can be done out of total n things. This is denoted by nCr which is equal to n!/r!(n-r)!.

Hope it helps...

Answer 2

Solution :

Given ,

$\frac{n!}{r!(n-r)!}$ ---( 1 )

i ) if n = 10 , r = 4 then

$\frac{10!}{4!(10-4)!}$

= $\frac{10.9.8.7.6!}{(4.3.2.1)(6!)}$

= $ 10×3×7 $

= $210$

ii ) If n = 15 , r = 15 then

$\frac{15!}{15!(15-15)!}$

= $\frac{1}{0!}$

= $1$

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