How many ways are there that no two students will have the same birth date in a class of size 60

I just started a probability course and i'm not sure if i should use a combination or permutation for this question: How many ways are there that no two students will have the same birth date in a class size of 60? Should i do 365C60 or 365P60? Thanks so much!

-Amanda

I would actually start from basics pick the first student, can choose from 365 unique bdays, the 2nd can choose from unique 364.. .. the 60 can choose from unique 365-59 so you get a total of 365*364*...*(365-59)

which does this look most like?

Then you should check what the question is exactly asking, can you assume the students are distinuigishable, which I think you can for this case

Yeah I should have thought about the distinguishable factor. combinations dont care about any particular order so they can have repeats of cases....as far as what ive learned about them...whereas permutations only count specific outcomes once [making the outcomes distinguishable] and yes, i see now that ishould use a permutation i had written down 365!/(365-60)! which is a permutation but i dont think i was clearly thinking about the difference between permutation and combination so i was afraid i was missing a factor in the denominator. this sounds right! thanks so much for the clarification! : ) by the way; next time ill write down what i have worked out so far in the original post.

-Amanda

I just started a probability course and i'm not sure if i should use a combination or permutation for this question: How many ways are there that no two students will have the same birth date in a class size of 60? Should i do 365C60 or 365P60? Thanks so much!

-Amanda

BBB

The first student can have any of 366 birth dates. The second student can have any of those 366 except the one taken by the first student. The third student can have any of those 366 except the ones taken by the first two students.

etc.

This is the classic birthday problem. There are really a bunch of ways. Think of the students as numbered 1 through 60. There are 366 (leap year birthday possible) days for 1's birthday. That leaves 365 days for 2's birthday, 364 for 3's birthday, etc. So in total there are $$P(366,307)=366\cdot365\cdot364\cdots307\approx 3.84\times10^{151}$$

total ways.