The number of possible options to get colored balls I have a bag with 5 white balls and 3 blue balls. We getting out 3 balls (without returning).
How many possible ways is to get at least 1 white ball? 
My question is why this is not the answer:
$${{5}\choose{1}}*{7 \choose 2}$$
 A: The number of ways of selecting $k$ of the five white balls and $3 - k$ of the three blue balls is 
$$\binom{5}{k}\binom{3}{3 - k}$$
Thus, the number of ways of selecting at least one white ball is 
$$\binom{5}{1}\binom{3}{2} + \binom{5}{2}\binom{3}{1} + \binom{5}{3}\binom{3}{0}$$
Alternatively, we could find the same total by subtracting the number of ways of selecting zero white balls from the total number of ways of selecting three of the eight available balls.  That yields
$$\binom{8}{3} - \binom{5}{0}\binom{3}{3}$$
Why did your calculation lead to an incorrect result?
You selected one of the five white balls and two of the seven other balls.  However, this counts those selections in which more than one white ball is selected more than once.  
Suppose you selected two particular white balls.  You counted each selection twice, once for each way you could have selected one of those two white balls as the white ball you selected from the five white balls while choosing the other white ball from the remaining seven balls.  
You counted each selection of three particular white balls three times, once for each of the three ways you could have selected one of those three white balls as the white ball you selected from the five white balls while selecting the other two selected white balls from the remaining seven balls.
Notice that 
$$\binom{5}{1}\binom{3}{2} + 2\binom{5}{2}\binom{3}{1} + 3\binom{5}{3}\binom{3}{0} = \binom{5}{1}\binom{7}{2}$$
