# 12 students in a class, how many ways these can take 4 different tests if 3 students are to take each test?

Can I approach the problem as 12-digit number with each digit having $4$ possible values and then $3$ digits must take $4$ values , so C($12$,$3$)*$4^3$ and how to do the rest part for remaining $12-3=9$ digits ?

How many options are there for which three take the first test? There are $12 \choose 3$ options. Then there are 9 remaining students. How many options are there for which three of these nine take the second test? There are $9 \choose 3$ options. Keep iterating this logic, and you'll get
$$\text{Total number of options} = {12 \choose 3} \cdot {9 \choose 3} \cdot {6 \choose 3} = 369,600$$
• I really appreciate your writing, but I haven't understood how you are dividing students into a group of three and then make them take each test. Doesn't the question specify that $3$ students are to take all $4$ tests. And we need to find about the remaining ? Commented Jun 4, 2015 at 18:12
• Sure. As I interpret the question, there are 12 students and four tests. Each test must be taken by exactly three students. Then, how many ways can we pick 3 students for the first test? $12 \choose 3$. How many ways are there to pick 3 students from the remaining 9 for the second test? $9 \choose 3$. And so forth. Do you think I'm misinterpreting the question? If so, please explain. Commented Jun 4, 2015 at 18:56