Question

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least two balls of each color?

Answer 1

Hint: This problem deals with permutations and combinations. But here a simple concept is used. Although this problem deals with combinations only. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$ \Rightarrow n! = n(n - 1)(n - 2).......1$
The no. of combinations of $n$ objects taken $r$ at a time is determined by the formula which is used:
$ \Rightarrow {}^n{c_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step-by-step answer:
The no. of red color balls in the box = 5
The no. of white color balls in the box = 6
We have to select 6 balls from the box so that there should be 2 balls at least of each color, which means that there can be more than two balls from each color.
Here selecting the no. of combinations in total 6 balls choosing balls from red and white colors, which have at least two from each color.
Considering the cases of selection as given below:
Selecting 4 red balls from 5 red balls and 2 white balls from 6 white balls which can be done in the following ways, as given below:
$ \Rightarrow {}^5{c_4} \times {}^6{c_2}$
Now selecting 3 red balls from 5 red balls and 3 white balls from 6 white balls which can be done in the following ways, as given below:
$ \Rightarrow {}^5{c_4} \times {}^6{c_3}$
Now selecting 2 red balls from 5 red balls and 4 white balls from 6 white balls which can be done in the following ways, as given below:
$ \Rightarrow {}^5{c_2} \times {}^6{c_4}$
So the total no. of ways that 6 balls be selected so that there are at least two balls of each color, is given by:
$ \Rightarrow {}^5{c_4} \times {}^6{c_2} + {}^5{c_4} \times {}^6{c_3} + {}^5{c_2} \times {}^6{c_4}$
\[ \Rightarrow 75 + 200 + 150 = 425\]

Final Answer: The no. of ways that 6 balls be selected so that there are at least two balls of each color is 425.

Note:
Please note that while solving this problem, we are asked to select 6 balls at least from each color, that means we cannot select less than 2 balls from each color. That is the reason we selected 3 and 3 from each color, or 4 and 2 from each color and 2 and 4 from each color, as already discussed there cannot be less than 2 balls from each color.