document.write( "Question 1196161: a box contains 4 red, 3 blue, 2white balls. In how many ways can we select 3 balls such that
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Algebra.Com's Answer #828922 by math_helper(2461) $\"\"$ $\"About$

You can put this solution on YOUR website!


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document.write( "a) 4*3*2 * 3! = 144 ways\r
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document.write( "b) 4*3*2 = 24 ways\r
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document.write( "c) 3*2*2*3 = 36 ways\r
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document.write( "d) BBW: 3*2*2*3 = 36   PLUS   BBR: 4*3*2*3 = 72,  so the total here 
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document.write( "is  36 + 72 =  108 ways\r
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document.write( "I will give you this table, you can complete the rest:\r
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document.write( "   Draw                                  Number of ways
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document.write( "   -----                                 --------------
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document.write( "   RWB (and the 5 other arrangments):      144 
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document.write( "   BBB                                       6
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document.write( "   WWW                                       0
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document.write( "   RRR                                      24
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document.write( "   RRB (and RBR, BRR)                      108
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document.write( "   RRW (RWR, WRR)                           72               
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document.write( "   RWW (WRW, RWW)                           24
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document.write( "   BBW (BWB, WBB)                           36
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document.write( "   RBB (BRB, BBR)                           72
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document.write( "   BWW (WBW, WWB)                           18\r
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document.write( "                          TOTAL            504                                        
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document.write( "Example:  RRW is computed as follows
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document.write( "         4 reds possible on first draw
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document.write( "       * 3 R on 2nd draw
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document.write( "       * 2 W on 3rd draw
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document.write( "       * 3  (b/c you can draw RRW or RWR or WRR and end up 2 R + 1 W)
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document.write( "      =======
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document.write( "        72   \r
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document.write( "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).\r
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