Question

g Four distinct numbers are to be selected from the list {−5, −4, −3, −2, −1, 1, 2, 3, 4}. In how many ways can this selection be done so that the product of the four numbers is positive?

Answer 1

Answer: 66 possible combinations.

Step-by-step explanation:

To have a positive product we have 3 situations.

The 4 numbers are positive:

if the "order" of the selection does not matter, then we have only one solution here:

1, 2, 3 and 4.

Second case, we have two negative numbers and two positive numbers.

Here we can use the fact that in a group of N objects, the number of different combinations of K objects (where K ≤ N) is:

$C = \frac{N!}{(N -K)!*K!}$

Here we have 5 negative numbers and we want to make groups of 2, so the possible combinations are:

$C = \frac{5!}{3!*2!} = \frac{5*4}{2*1} = 2*5 = 10$

And we have exactly the same for the other two positive numbers, but in this case we have N = 4 and  K = 2.

$C = \frac{4!}{2!*2!} = 6$

The total number of combinations is the product of those two:

C = 10*6 = 60 combinations

Now, the last option is that the 4 numbers are negative numbers, so here we have 5 negative numbers and we want to make groups of 4.

$C = \frac{5!}{1!*4!} = 5$

So in total, we have: 1 + 60 + 6 = 66 possible combinations.