Probability of picking up one ball of each color A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Let P be the probability that among the balls drawn there is at least one ball of each color. Find 455 * P.
My approach:
I reasoned that because we want at least 1 ball of each color, I could proceed like:
$\frac{6*5*4*12}{15 x 14x13x12}\quad X \quad455$
So first I can choose between the six red balls, then the 5 black balls and finally the 4 white balls. Then, since we already have 1 of each color, we choose from the remaining 12.
So with this, I got 20 as the answer. However, the answer is supposed to be 240. I understood the solution that was given, however, I wanted to know where my solution  was flawed and how I could improve it.
Any help will be appreciated.
 A: Personally, I find this method much simpler. 
Red-white-black need to be drawn in 2-1-1 ,1-2-1 or 1-1-2 pattern, so simply apply the hypergeometric formula:
$$455\times \dfrac{{6\choose 2}{4\choose 1}{5\choose1}+{6\choose 1}{4\choose2}{5\choose 1} +{6\choose1}{4\choose 1}{5\choose 2}}{15\choose 4} = 240$$
NOTE
Make a note that whenever you try to "fractionate" the combinations, by fulfilling the conditions in instalments the way you did, you will always overcount. 
A: Use inclusion/exclusion principle:
The number of ways to draw at least one ball of each color:


*

*Include the number of ways to draw any balls: $\binom{6+5+4}{4}=1365$

*Exclude the number of ways to draw only RB:   $\binom{6+5  }{4}= 330$

*Exclude the number of ways to draw only RW:   $\binom{6+4  }{4}= 210$

*Exclude the number of ways to draw only BW:   $\binom{5+4  }{4}= 126$

*Include the number of ways to draw only R:    $\binom{6    }{4}=  15$

*Include the number of ways to draw only B:    $\binom{5    }{4}=   5$

*Include the number of ways to draw only W:    $\binom{4    }{4}=   1$


Hence the probability to draw at least one ball of each color:
$$\frac{1365-330-210-126+15+5+1}{1365}=\frac{48}{91}$$
Multiplying this by $455$ will give you the desired result of $240$.
A: We can still proceed ahead from the way you counted $ 6*5*4*12 = 1440 $ & subtract the duplicates. For example duplicates for:
Red balls $$ \binom{6}{2} * 5 *4 = 300 $$
for Black balls $$ \binom{5}{2} * 6 *4 = 240 $$
for White balls $$ \binom{4}{2} * 6 *5 = 180 $$
so net $1440-300-240-180 = 720$ out of a total of $\binom {15}{4} = 1365$ giving $ P= 720/1365$ and $455* P =240.$
