A company could advertise its new product in 5 magazines, 4 newspapers, and 3 television channels. But in a later move, it decided to give advertisements in only 2 of the magazines, one of the newspapers and one of the TV channel. In how many ways can they advertise?

The company has several options for advertising.

First, let's consider the magazines. The company wants to advertise in 2 out of 5 magazines. This is a combination problem, because the order in which the magazines are chosen does not matter. The number of ways to choose 2 magazines out of 5 is given by the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of options, k is the number of options to choose, and "!" denotes factorial.

So, the number of ways to choose 2 magazines out of 5 is C(5, 2) = 5! / [2!(5-2)!] = 10.

Next, let's consider the newspapers. The company wants to advertise in 1 out of 4 newspapers. Again, this is a combination problem. The number of ways to choose 1 newspaper out of 4 is C(4, 1) = 4! / [1!(4-1)!] = 4.

Finally, let's consider the TV channels. The company wants to advertise in 1 out of 3 TV channels. This is also a combination problem. The number of ways to choose 1 TV channel out of 3 is C(3, 1) = 3! / [1!(3-1)!] = 3.

To find the total number of ways the company can advertise, we multiply the number of options for each type of media, because the choices are independent of each other.

So, the total number of ways the company can advertise is 10 (magazines) * 4 (newspapers) * 3 (TV channels) = 120 ways.