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 ?


1 Answer 1


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 should add that I'm assuming each student takes exactly one test.

  • $\begingroup$ 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 ? $\endgroup$ Commented Jun 4, 2015 at 18:12
  • $\begingroup$ do you mind explaining ?please $\endgroup$ Commented Jun 4, 2015 at 18:18
  • $\begingroup$ 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. $\endgroup$
    – Shane
    Commented Jun 4, 2015 at 18:56
  • $\begingroup$ There is no doubt about the solution you have provided, considering the way you have interpreted the question. But Iam still wondering about the question itself. What do we need to infer from it ? $\endgroup$ Commented Jun 4, 2015 at 19:04
  • $\begingroup$ Haha - I'm afraid only you, as the poser of the question, can really divine the correct interpretation of the question. It's a little ambiguous, but my interpretation seems about as straight-forward as possible. $\endgroup$
    – Shane
    Commented Jun 4, 2015 at 19:05

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