Science:Math Exam Resources/Courses/MATH110/December 2013/Question Section 001 10 (a)

MATH110 December 2013

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Question Section 001 10 (a)
How many ways can $7$ graduate students be assigned to teach the $4$ sections of MATH110 next year if each graduate student can be assigned to at most one section? For this question, a numerical answer is required as a final answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Here it matters which instructor teaches which course. How many choice of instructors are there for the first course? For the second course? Do you need to add or multiply these numbers?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We have $4$ slots to fill and $7$ people to fill them. So that means, we have $7$ choices for the first slot, $6$ for the second, $5$ for the third and $4$ for the fourth. Since the number of choices for each slot does not dependent on the choice of the previous spots we need to multiply these numbers. So the total number of ways to fill the slots is $7\cdot 6\cdot 5\cdot 4={\frac {7!}{3!}}=840$. There are $840$ ways to fill the slots.

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