see below for solution

The answer posted was 85. Can anyone explain how to arrive at 85 or confirm that this is not the correct answer?

When they say "how many ways," apparently what they mean is that each pick is unique, so that if you choose red ball number one on your first draw, for example, it's different than choosing red ball number two on the second draw, even though you have the same color at the end of each draw. in other words, order doesn't matter, but each ball is unique, in which case, you must use rules of permutations and combinations.

Specifically, when we want to find the number of combinations of a set of size 3 without repeated balls that can be made from the ten balls in the box, and order doesn't matter (RB is the same as BR), we must first find out all the possible combinations of 3 that can be taken from 10 (10_C_3). Then we must find all the ways that three colors in those groups of size 3 can be arranged: 3 x 2 x 1 = 3! = 6. Thus the total number of permutations of size 3 taken from a set of size 10 is equal to 3! times the total number of combinations of size 3 taken from a set of size 10: 10_P_3 = 7! x 10_C3.

When we divide both sides of this equation by 3! we see that the total number of combinations of size 3 taken from a set of size 10 is equal to the number of permutations of size 3 taken from a set of size 10 divided by 7!. This makes it possible to write a formula for finding 10_C_4: 7

10_P_3 10! 10!
10_C_3 = -------- = ------- = ----------
4! 3! x 7! 3!(10-3)!


10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= --------------------------------------
3 x 2 x 1 (7 X 6 x 5 x 4 x 3 x 2 x 1)


10 x 9 x 8 720
= -------------- = ------ = 120
3 x 2 x 1 6

How you get to 85, is you must now subtract out all the combinations of three balls in the above group that do not include red balls, or rather simply figure the number of combinations of the white and black balls or:

7! 5040
----------- = -------- = 35
(3! X 4!) 144


120 (all possible combinations of three) - 35 ( white and black only combinations of three) = 85 combinations where at least one is red.

Hope this helps.

possibilities
1 red, 2 others
2red, 1 other
3 red, no others

1 red, 2 others=3C1*7C2=63
2red, 1 other=3C2*7=21
3 red, no others=1
add all of them. Answer is 85

No. of ways 3 balls can be selected such that atleast one is red =
No. ways 3 balls can be selected from the whole set - No. of ways 3 balls can be selected from color other than red

Therefore Ans = 10C3 - 7C3 =( 10 X 9 X 8/3!) - (7 X 6 X 5/3!) = 120-35=85