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combinatory

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From a group of 10 professors how many ways can a committee of 5 members can be formed so that atleast one of the professor A and professor B will be included ?
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Best method to solve  at least problems in exams is

No of ways that at least one is present = (total no of ways )- (no of ways both are not present) = 10C5 - 8C5 = 252-56 = 196

Another approach

No of ways that at least one is present = (first professor is present)+(second professor is present)- (both are present)

= 9C4+9C4-8C3 = 196

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