Question

Show that the no of combinations of 6 dissimilar objects taken 3 at a time is 20 and for the same the no of permutations is 120

Answer 1

Answer:

Step-by-step explanation:

$Number\:of\:combinations\:required =n{C_r}$

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

$=\frac{6!}{3!(6-3)!}$

$=\frac{6!}{3!.3!}$

$=\frac{1.2.3.4.5.6}{1.2.3.1.2.3}$

$=20$

$Number\:of\:permutations\: required =n{P_r}$

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

$=\frac{6!}{(6-3)!}$

$=\frac{6!}{3!}$

$=\frac{1.2.3.4.5.6}{1.2.3}$

$=120$