Probability: Permutations/Combinations

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The discussion revolves around calculating probabilities related to students enrolling in sections of a course. The sample space for three students choosing from four sections is clarified as 4^3, equating to 64 possible outcomes. The probability of all students selecting the same section is derived as 3/64, while the probability of them choosing different sections is calculated as 6/64. Additionally, the probability of none selecting section 1 is determined to be 6/64. The conversation emphasizes the importance of clearly presenting homework problems to facilitate understanding and assistance.
catbearbig
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I'm having a hard time understanding how to derive the # ways to do things.

(a) A course has 4 sections with no limit on how many can enrol in each section. 3 students each pick a section at random.

(i) Specify Sample Space S (64?)
(ii)Find the probability that they all end up in the same section (3C1 /64 = 3/64??)
(iii) Find the probability that they all end up in different sections (3P3/64 = 6/64 ??)
(iv) Find the probability that nobody picks section 1 ( 3C2 * 2! / 64 = 6/64??)

(b) Repeat (a) in the case where there are n sections and s students.

Can someone please help me derive their answers of how they get the # of ways to do each. Thanks a lot!
 
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I suggest, first, that you check the top thread on how to present a homework problem. Or look at the standard layout used by most others.

For your probability answer, don't forget there are 4 sections, not 3. And a little more explanation of your thinking would help others to help you.
 

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