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

Answer: (A)

Case-I : when exactly one box provides four balls ( $3R1B$ or $2R2B$ )
Number of ways in this case ${}^5{C}_4{\left({}^3{C}_1\times {}^2{C}_1\right)}^3\times 4$
Case-II : when exactly two boxes provide three balls ( $2R1B$ or $1R2B$ ) each
Number of ways in this case ${\left({}^5{C}_3-1\right)}^2{\left({}^3{C}_1\times {}^2{C}_1\right)}^2\times 6$
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.