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 at least 1 blue can and 1 red can remain in the refrigerator.
Answers
Answered by
1
Answer: 455
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.
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.
Answered by
0
7c3*5c5+7c4*5c4+7c5*5c3+7c6*5c2+7c7*5c1
= 7*5 + 35*5 + 21*10 + 70 + 5
= 35 + 175 + 210 + 75
=495
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