Set S = {-1, 0, 1, 2, 3, 4, 5}

We can select 3 integers from these 7 integers in 7C3 ways
7C3 = (7)(6)(5)/(3)(2)(1) = 35
So, it SEEMS like there might be 35 different possible products.

HOWEVER, among these 35 possible products, we have some duplication.
For example, if our 3 selected integers are -1, 0 and 2, the product = 0
And, if our 3 selected integers are 0, 3 and 5, the product = 0
And, if our 3 selected integers are 0, 1 and 4, the product = 0
In fact, each time 0 is among the three selected integers, the product will be 0.

So, in how many different ways will 0 be among the three selected integers?
Let's find out.

We'll first choose 0 to be one the three selected integers.
Now choose 2 more integers from {-1, 1, 2, 3, 4, 5}
We can select 2 integers (to join our already-selected 0) in 6C2 ways
6C2 = (6)(5)/(2)(1) = 15
So, there are 15 different ways to get a product of 0.
We can only count 0 as a product ONE TIME.

So among the 35 outcomes we calculated earlier, we counted the product of zero 15 times (when we should count in ONE TIME).

So, we must subtract 14 from our original 35 outcomes to get 21

Answer: 21

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The number of triplets which contains 0 is 15, thus there are 14 copies. The number of unique results is 35-14 = 21.

Can somebody please explain this portion ?

We're looking for the number of DIFFERENT products (of the 3 selected numbers)
There are 15 ways to get a product of zero.
Since we can only count the product of zero 1 time, we must subtract the other 14 duplicates.

Does that help?

Cheers,
Brent

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