# Combinations problem

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I am confused why the answer used the combination to be multiplied by the probability but I will try to go through the whole steps with my explanation and would appreciate your help in checking my understanding here. Regards

Q) 5 students are in a class. What is the probability that exactly 2 have the same birthday?

A) First of all we can find the permutation (p1) for the all the outcomes of having exactly 2 with the same birthday and that is 365x1x364x363x362

Then to find the permutation (p2) representing all the possible outcomes which is the sample space 365^5

Now the probability is (p1/p2 = p) but this probability represents having exactly 2 students with the same birthday and knowing which students they are when selecting. But we are going to select randomly and do not know which students are having the same birthday. We can have 5C2 (=10 different combinations) so px10 will give us the probability for selecting at random and still having 2 with the same birthday.

That is my understanding to this and your thoughts are welcome.

Many thanks in advance.

Q) 5 students are in a class. What is the probability that exactly 2 have the same birthday?

A) First of all we can find the permutation (p1) for the all the outcomes of having exactly 2 with the same birthday and that is 365x1x364x363x362

Then to find the permutation (p2) representing all the possible outcomes which is the sample space 365^5

Now the probability is (p1/p2 = p) but this probability represents having exactly 2 students with the same birthday and knowing which students they are when selecting. But we are going to select randomly and do not know which students are having the same birthday. We can have 5C2 (=10 different combinations) so px10 will give us the probability for selecting at random and still having 2 with the same birthday.

That is my understanding to this and your thoughts are welcome.

Many thanks in advance.

Last edited by DanielDaniels; 4 weeks ago

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#2

Which bit are you confused about? Why you multiply by 10 or ...

Last edited by mqb2766; 4 weeks ago

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(Original post by

Which bit are you confused about? Why you multiply by 10 or ...

**mqb2766**)Which bit are you confused about? Why you multiply by 10 or ...

Last edited by DanielDaniels; 4 weeks ago

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#4

There are obviously two parts to the problem

1) The probability that two share, and the other 3 are different

2) Counting (summing) all possible share combinations

Im assuming you can do 1), for the second one there are 20 = 5*4 = 5P2 ordered combinatations. However, (Fred, Jane) is the same as (Jane, Fred) so the order is unimportant so there are 5*4/2 = 5C2 unordered combinations.

https://www.mathsisfun.com/combinato...mutations.html

is a bit long, but a pretty good read.

1) The probability that two share, and the other 3 are different

2) Counting (summing) all possible share combinations

Im assuming you can do 1), for the second one there are 20 = 5*4 = 5P2 ordered combinatations. However, (Fred, Jane) is the same as (Jane, Fred) so the order is unimportant so there are 5*4/2 = 5C2 unordered combinations.

https://www.mathsisfun.com/combinato...mutations.html

is a bit long, but a pretty good read.

Last edited by mqb2766; 4 weeks ago

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(Original post by

There are obviously two parts to the problem

1) The probability that two share, and the other 3 are different

2) Counting (summing) all possible share combinations

Im assuming you can do 1), for the second one there are 20 = 5*4 = 5P2 ordered combinatations. However, (Fred, Jane) is the same as (Jane, Fred) so the order is unimportant so there are 5*4/2 = 5C2 unordered combinations.

https://www.mathsisfun.com/combinato...mutations.html

is a bit long, but a pretty good read.

**mqb2766**)There are obviously two parts to the problem

1) The probability that two share, and the other 3 are different

2) Counting (summing) all possible share combinations

Im assuming you can do 1), for the second one there are 20 = 5*4 = 5P2 ordered combinatations. However, (Fred, Jane) is the same as (Jane, Fred) so the order is unimportant so there are 5*4/2 = 5C2 unordered combinations.

https://www.mathsisfun.com/combinato...mutations.html

is a bit long, but a pretty good read.

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#6

(Original post by

Thank you. I do get the reason why it is 10 but my problem is that why actually we needed this second part of multiplying the combinations?

**DanielDaniels**)Thank you. I do get the reason why it is 10 but my problem is that why actually we needed this second part of multiplying the combinations?

Last edited by mqb2766; 4 weeks ago

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(Original post by

Why its 10*p? Each outcome has probabiltiy p of occurring so you sum them as they're mutually exclusive. So summing 10 p's = 10p

**mqb2766**)Why its 10*p? Each outcome has probabiltiy p of occurring so you sum them as they're mutually exclusive. So summing 10 p's = 10p

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#8

(Original post by

Now I get it. Many thanks for your help here.

**DanielDaniels**)Now I get it. Many thanks for your help here.

p(pair1 OR pair2 OR pair3 ...) = p(pair1) + ... + p(pair10) - 0 - 0 - ... - 0

The 0's are the p(pair1 AND pair2) which are 0 because the pairs are mutually exclusive.

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(Original post by

I edited the previous post slightly, but to be clear, you're calculating

p(pair1 OR pair2 OR pair3 ...) = p(pair1) + ... + p(pair10) - 0 - 0 - ... - 0

The 0's are the p(pair1 AND pair2) which are 0 because the pairs are mutually exclusive.

**mqb2766**)I edited the previous post slightly, but to be clear, you're calculating

p(pair1 OR pair2 OR pair3 ...) = p(pair1) + ... + p(pair10) - 0 - 0 - ... - 0

The 0's are the p(pair1 AND pair2) which are 0 because the pairs are mutually exclusive.

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