If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator.
455
255
555
355
Explanation:
Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4)
(6,2) =>7C6×5C27C6×5C2 => 710 = 70
(5,3) =>7C5×5C37C5×5C3 => 21 x 10 = 210
(4,4) =>7C4×5C47C4×5C4 => 35 x 5 = 175
70 + 210 + 175 = 455.