How Should Birthdays Be Counted for a Class of 60?

  • Thread starter Thread starter kirkulator
  • Start date Start date
  • Tags Tags
    Probability
kirkulator
Messages
32
Reaction score
0
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
 
Physics news on Phys.org
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 don't care about any particular order so they can have repeats of cases...as far as what I've 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 don't 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
 
Last edited:
kirkulator said:
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

Don't forget that a few million people were born on Feb 29th...

BBB
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top