- Home
- Standard 11
- Mathematics
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?
$21816$
$85536$
$12096$
$156816$
Solution

$Case-I$ : when exactly one box provides four balls ( $3 R 1 B$ or $2 R 2 B$ )
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 ( $2 R 1 B$ or $1 R 2 B$ ) 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.