If a refrigerator contains 12 cans such that 7 are blue and 5 are red, in how…

Concept: When a selection must satisfy a constraint (here, keeping at least one can of each colour behind), it is usually faster to count the complement — find the total number of unrestricted selections and subtract the cases that violate the constraint — rather than building every valid case directly. The number of ways to choose r objects out of n is nCr = n!/(r!(n−r)!).

1. Removing 8 cans out of 12 is equivalent to choosing which 4 cans stay behind, since the total pool of 12 is fixed.

2. The condition “at least 1 blue and 1 red can remain” fails only when the 4 remaining cans are all one colour — either all 4 remaining are blue, or all 4 remaining are red.

3. Total ways to choose any 4 cans to remain out of 12 = 12C4 = 495.

4. Ways in which all 4 remaining cans are blue (choosing 4 of the 7 blue cans) = 7C4 = 35.

5. Ways in which all 4 remaining cans are red (choosing 4 of the 5 red cans) = 5C4 = 5.

6. Subtracting both disallowed cases from the total: 495 − 35 − 5 = 455.

Cross-check: This can also be verified by directly summing every way the remaining 4 cans can include both colours — splits of 3 blue+1 red, 2 blue+2 red, and 1 blue+3 red among the remaining cans: 7C3×5C1 + 7C2×5C2 + 7C1×5C3 = 35×5 + 21×10 + 7×10 = 175 + 210 + 70 = 455, matching the complement-method result.

Result: The number of valid ways is 455.