SOLUTION: a box contains 4 red, 3 blue, 2white balls. In how many ways can we select 3 balls such that a). they are different colors? b). they are all red c). two are blue and one is wh

Algebra ->  Permutations -> SOLUTION: a box contains 4 red, 3 blue, 2white balls. In how many ways can we select 3 balls such that a). they are different colors? b). they are all red c). two are blue and one is wh      Log On


   



Question 1196161: a box contains 4 red, 3 blue, 2white balls. In how many ways can we select 3 balls such that
a). they are different colors?
b). they are all red
c). two are blue and one is white?
d). exactly 2 are blue?
e). non is white
f). at least one white?

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

a) 4*3*2 * 3! = 144 ways
b) 4*3*2 = 24 ways
c) 3*2*2*3 = 36 ways
d) BBW: 3*2*2*3 = 36 PLUS BBR: 4*3*2*3 = 72, so the total here
is 36 + 72 = 108 ways
I will give you this table, you can complete the rest:
Draw Number of ways
----- --------------
RWB (and the 5 other arrangments): 144
BBB 6
WWW 0
RRR 24
RRB (and RBR, BRR) 108
RRW (RWR, WRR) 72
RWW (WRW, RWW) 24
BBW (BWB, WBB) 36
RBB (BRB, BBR) 72
BWW (WBW, WWB) 18
TOTAL 504

Example: RRW is computed as follows
4 reds possible on first draw
* 3 R on 2nd draw
* 2 W on 3rd draw
* 3 (b/c you can draw RRW or RWR or WRR and end up 2 R + 1 W)
=======
72
If you need probabilities, divide each number of ways by 504 (=9*8*7, this comes from the total number of balls in the box at first draw, 2nd draw, 3rd draw).