Here is how:

There are totally 6 senior staff members out of which 3 have to be picked. This is a combination problem. Remember: order doesn't matter here. If order did matter, it would be a permutation problem and your answer would be right.
The number of ways this can be done is 6C3 .i.e (6*5*4) / (1*2*3) = 20 ways

There are four junior staff members and 1 has to be choosed from them. Again a combination problem. The number of ways is 4C1 = (4) / (1) = 4

Considering both as independent events, the number of ways in which 3 senior members AND 1 junior member can be selected from 6 senior and 4 junior staff members is 20 * 4 = 80 ways.

ufdan wrote:If there are 6 senior staffers and 4 junior staffers and a committee consisting of 3 seniors and 1 junior is to be formed, how many possible committees can be formed?

I tried solving by multiplying the number of possible senior members 6x5x4 by the number of possible junior members 4. I know this is wrong but I am at a loss on how to approach this now...

You did all well but forgot to divide by the "repeated seniors" as the order doesn't matter.

SSSJ/ 3!