An urn contains 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls:
(a) If 2 balls are white and 1 is red?

(b) If all 3 balls are white?

(c) If all 3 balls are red?

We are given an urn containing 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls if 2 balls are white and 1 is red

We know that we have a total of 7 white ball and 3 red balls.The ways to draw 2 white ball and 1 red ball is :

$C(7,2)\times C(3,1)=\frac{7!}{2!5!}\times \frac{3!}{1!2!}$

$\frac{7\times 6\times 5!}{2\times 1\times 5!}\times \frac{3\times 2!}{1\times 2!}=\frac{7\times 6}{2}\times \frac{3}{1}$

$7\times 3\times 3=63$ways

We are given an urn containing 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls if all 3 balls are white?

We know that we have a total of 7 white ball and 3 red balls.The ways to draw 3 white balls :

$C(7,3)=\frac{7!}{3!}$

$\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}=\frac{7\times 6\times 5}{3\times 2\times 1}$

$7\times 5=35$ways

We are given an urn containing 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls if all 3 balls are red?

We know that we have a total of 7 white ball and 3 red balls.The ways to draw 3 red balls :

$C(3,3)=\frac{3!}{3!\times 0!}$

$\frac{3!}{3!}=1$way