Question

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

• A. 21816
• B. 85536
• C. 12096
• D. 156816

Answer 1

The Correct Option is A

Case-I : when exactly one box provides four balls ( 3R1B or 2R2B )
Number of ways in this case \(5C_4 \times (3C_1 \times 2C_1)^3 \times 4\)
Case-II : when exactly two boxes provide three balls ( 2R1B or 1R2B ) each
Number of ways in this case \((5C_3 - 1)^2 \times (3C_1 \times 2C_1)^2 \times 6\)
Required number of ways =21816
Language ambiguity : If we consider at least one red ball and exactly one blue ball, then required number of ways is 9504 . None of the option is correct.
The correct option is (A): 21816

Answer 2

There are 4 bags and every bag has 3 Red and 2 Blue balls
\(\begin{matrix} &A&B&C&D \\ Case1&4Balls&2Balls&2Balls&2Balls \\ Case2&3Balls&3Balls&2Balls&2Balls\end{matrix}\)
Among four bags, one bag is selected to take out 4 balls. Determine the number of ways to choose one bag from these 4 bags.\(^4C_1\)
Ways of taking 4 balls:
1. 3R and 1B balls,  \(^3C_3 \times\ ^2C_1\)
2. 2R and 2B balls,  \(^3C_2 \times\ ^2C_2\)
Two balls are taken from remaining three bags,  \(^3C_1 \times\ ^2C_1\) ways and there are 3 bags so, \([\ ^3C_1 \times\ ^2C_1]^3\)
Total number of ways of chosing 10 balls from these four bags \(^4C_1[\ ^3C_3 \times\ ^2C_1+\ ^3C_2 \times\ ^2C_2] \times [\ ^3C_1 \times\ ^2C_1]^3\)  Eq-1

Among four bags, two bag is selected to take out 3 balls. Determine the number of ways to choose two bag from these 4 bags.\(^4C_2\)
Ways of taking 3 balls are:
1. 2R and 1B balls, \(^3C_2 \times\ ^2C_1\)
2. 1R and 2B balls, \(^3C_1 \times\ ^2C_2\)
So, for two bags three balls can be taken in \([^3C_2 \times\ ^2C_1+^3C_1 \times\ ^2C_2]^2\) ways.
Now, two balls are taken from remaining two bags, \(^3C_1 \times\ ^2C_1\) and there are two bags so, \([^3C_1 \times\ ^2C_1]^2\)
Total number of ways of chosing 10 balls from these four bags: \(^4C_2[^3C_2 \times\ ^2C_1+^3C_1 \times\ ^2C_2]^2 \times [^3C_1 \times\ ^2C_1]^2\)  Eq-2

Now add both the equation 1 and 2
\(^4C_1[\ ^3C_3 \times\ ^2C_1+\ ^3C_2 \times\ ^2C_2] \times [\ ^3C_1 \times\ ^2C_1]^3+^4C_2[^3C_2 \times\ ^2C_1+^3C_1 \times\ ^2C_2]^2 \times [^3C_1 \times\ ^2C_1]^2\)
\(=4[5][6]^3+6[9]^2[6]^2\)
\(=4320+17496\)
\(=21816\)

So, the correct option is (A): 21816