Aclass has 12 students, of which 2 are seniors. how many committees of size 5 can be selected if at least one member of each committee must be a senior?

The concepts you have to learn about in here are permutations and computations. They are quite the same, but they only differ in technicalities. Permutations are arrangements where order matters, otherwise, the arrangements are called combinations. For example, if you want to choose 3 winners out of the five contestants, that is a combination problem. But if you want to choose the top 3 winners out of the five contestants, this time, it is a permutation problem because order matters by ranking.

Now, choosing members of the committee is assumed to be random, if not mentioned. So, we apply the equation for combination

n!/r!(n-r)!
where
n is the total number of species per kind
r is the number of species you pick out of the total

So, out of the 12 students, 2 are seniors and 10 are non-seniors. Now, you have to find the number of ways if r is at least 1. So that means r could be 1 or 2. Therefore, the solution would be:

(10C4+ 2C1) + (10C3 + 2C2)

10C4 means 4 committee students out of the 10 non-senior students. Note that the right side numbers after C must equal to 5 because the committee must contain 5 members. The left side numbers before C signify the total pool of species to choose from. Applying the equation

10!/4!(10 - 4)! + 2!/1!(2-1)! + 10!/3!(10-3)! + 2!/2!(2-2)!
= 333

Therefore, there are a total of 333 ways.