Question
There are nine distinct numbers of which five numbers are positive and four numbers are negative. Three numbers are chosen at random and the product of these numbers is found. How many of these products are positive?
Answer: Option A
:
A
For product of three numbers to be positive it must have either 0 or 2 negative numbers.
Case 1: 0 negative numbers.
3 positive numbers can be selected out of 5 in 5C3 ways = 10 ways.
Case 2: 2 negative numbers
2 negative numbers can be selected out of 4 in 4C2 ways = 6 ways.
1 positive number can be selected out of 5 in 5C1 ways = 5 ways.
Hence this combination can be selected in a total of (6×5)=30 ways.
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:
A
For product of three numbers to be positive it must have either 0 or 2 negative numbers.
Case 1: 0 negative numbers.
3 positive numbers can be selected out of 5 in 5C3 ways = 10 ways.
Case 2: 2 negative numbers
2 negative numbers can be selected out of 4 in 4C2 ways = 6 ways.
1 positive number can be selected out of 5 in 5C1 ways = 5 ways.
Hence this combination can be selected in a total of (6×5)=30 ways.
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