A bag has (5) distinct red and (7) distinct black balls. (4) balls are to be selected such that not all are of the same color. How many selections are possible?

Correct answer: (455)

The direct answer is option C, 455 selections. Since balls are distinct, a selection means choosing objects without caring about order. There are 12 balls altogether, so the total number of ways to choose 4 is \(^{12}C_4=495\). The condition says the selected balls must not all have one colour. The unwanted cases are: four red balls, counted by \(^{5}C_4=5\), and four black balls, counted by \(^{7}C_4=35\). These two cases cannot overlap. Therefore valid selections are \(495-5-35=455\). Option A, 410, subtracts too many or uses an incorrect count. Option B, 430, is also not the result of the required subtraction. Option C, 455, correctly removes both all-red and all-black selections. Option D, 475, does not remove the two forbidden groups correctly. Remember: for “not all of one type”, count everything and subtract every completely single-type case.