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
486
475
445
Explanation:
Possible ways to draw 8 balls from the refrigerator which contains atleast 1 blue and 1 red can
after the drawing are (6,2) (5,3) (4,4).
For (6, 2) $\Rightarrow$ 7c6 $\times$ 5c2 $\Rightarrow$ 7 $\times$ 10=70
For (5, 3) $\Rightarrow$ 7c5 $\times$ 5c3 $\Rightarrow$ 21$\times$ 10=210
For (4, 4) $\Rightarrow$ 7c4 $\times$ 5c4 $\Rightarrow$ 35 $\times$ 5=175
So Total ways = 70+210+175=455
Possible ways to draw 8 balls from the refrigerator which contains atleast 1 blue and 1 red can
after the drawing are (6,2) (5,3) (4,4).
For (6, 2) $\Rightarrow$ 7c6 $\times$ 5c2 $\Rightarrow$ 7 $\times$ 10=70
For (5, 3) $\Rightarrow$ 7c5 $\times$ 5c3 $\Rightarrow$ 21$\times$ 10=210
For (4, 4) $\Rightarrow$ 7c4 $\times$ 5c4 $\Rightarrow$ 35 $\times$ 5=175
So Total ways = 70+210+175=455