Question

Assuming the balls to be identical except for difference in colours, the number of ways in which
one or more balls can be selected from 10 white, 9 green and 7 black balls is :

Answer 1

not sure of a general formula here...but there is a pattern
your question is vague, so im assuming you want to know how many different color arrangements for selecting n balls, n=1,2,3...

n=1, 3 ways --> W or G or B
n=2 , 6 ways --> 2W or 2G or 2B or WG or WB or GB
n=3, 10 ways -->3W or 3B or 3G or 2W1G or 2W1B or 2G1W or 2G1B or 2B1W or 2B1G or WGB

and so on..
n=4, 15 ways

Answer 2

The answer shuld be 879

Answer 3

oh so you have to add them all up all the way to n=26

well there is symmetry, so n=0 is same as n=26 , n=1 same as n=25 and so on ...
as long as n is not greater than the number of possible color balls, then pattern is the num of ways equals (n+2) choose 2

\[\sum_{1}^{7}\left(\begin{matrix}n+2 \\ 2\end{matrix}\right) = 119\]
then figure out n=8 to 12 , add to 1198 , double it then add n=13

Answer 4

@experimentX @satellite73

Answer 5

@mukushla @siddhantsharan

Answer 6

11*10*8 - 1= 879 should be the answer.
I think.
Can you check?

Answer 7

Yes...u r correct...but hw did u do it...

Answer 8

We can select:
No white ball....1
One white ball....2
Two white balls.....3
.
.
.
.
10 white balls. .....11 => Total 11 ways to choose.
Similarly 10 ways for green and 8 ways for black.
However you will have to subtract the one case where no balls are chosen.
Therefore total=
(Ways to choose white)*(ways to choose black)*(ways to choose green) - 1
= 11*10*8 -1