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

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) (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).



 

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