We have 4 cases.

Case 1- 0 white balls:
So we have to choose 3 balls from 5 non-white balls. 5choose3 = 10

Case 2 - 1 white ball:
Since 1 will be white, we have to choose 2 balls from 5 non-white balls. 5choose2 = 10

Case 3 - 2 white balls
Since 2 will be white, we have to choose 1 ball from 5 non-white balls. 5choose1 = 1

Case 4 - 3 white balls
There is only 1 way to choose 3 white balls from a 3 white balls

Adding all 4 cases we get
10 + 10 + 5 + 1
= 26

Cheers.

maihuna wrote:In a box there are eight balls of which there are three white balls and five are of other colors all different. In how many ways three balls can be taken in a single draw:

13
26
52
60
70

"Case 2 - 1 white ball:
Since 1 will be white, we have to choose 2 balls from 5 non-white balls. 5choose2 = 10 "

ankit..a doubt

the numbe rof ways of selecting 1 whiteball and 2 other balls from 3 white balls and 5 other balls of different colours is

3c1 * 5c2 isnt it?

u have just used 5c2 without multiplying it with 3c1...

60

xcusemeplz2009 wrote:60

how?

The reason we do not multiply is because we do not differentiate between the different white balls.

The combination was:
Red, Blue,Green, Yellow and White1
Red, Blue,Green, Yellow and White2
Red, Blue,Green, Yellow and White3

if the white1, white2 and white3 are identical (say we called them all just white), then all the above combinations will be same. Hence we do not consider them as possible options.

Since the question asks how many ways..i'd assume it has how many different ways.

Hope it helps.

winnerhere wrote:"Case 2 - 1 white ball:
Since 1 will be white, we have to choose 2 balls from 5 non-white balls. 5choose2 = 10 "

ankit..a doubt

the numbe rof ways of selecting 1 whiteball and 2 other balls from 3 white balls and 5 other balls of different colours is

3c1 * 5c2 isnt it?

u have just used 5c2 without multiplying it with 3c1...