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): 10=2+2+2+4=2+2+3+3 In the first case the number of ways = ^4 C_1 * ( ^3 C_1 * ^2 C_1)^3 * ^5 C_4=4320 In the second case the number of ways = ^4 C_2 * ( ^3 C_1 * ^2 C_1)^2 * ( ^5 C_3-1)^2=6.6^2 x 9^2=216 x 81=17496 Total number of ways =17496+4320=21816

Consider 4 boxes, where each box contains 3 red balls a... | Permutations and Combinations (Mathematics)

(a):

10=2+2+2+4=2+2+3+3

In the first case the number of ways

={ }^4 C_1 \cdot\left({ }^3 C_1 \cdot{ }^2 C_1\right)^3 \cdot{ }^5 C_4=4320

In the second case the number of ways

={ }^4 C_2 \cdot\left({ }^3 C_1 \cdot{ }^2 C_1\right)^2 \cdot\left({ }^5 C_3-1\right)^2=6.6^2 \times 9^2=216 \times 81=17496

Total number of ways

=17496+4320=21816

Explanation

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

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