Hi, there are two problems I can't really solve yet they don't seem that difficult. The two of them seem pretty related to me, I think there's something I'm not getting I'll detail my attempts at solving but any help especially with the steps to the solution would be really appreciated as I have the answers, these are not homework or anything like that just me practicing.(adsbygoogle = window.adsbygoogle || []).push({});

The first one is: Consider 4 digits, what are the probabilities of getting 1 different digit, 2 of them, 3 of them or 4.

I have no problem with 1 digit, it is 10/1000. The third I can do also and the 4th, it is especially with two different digits that I struggle.

So I thought the two digits which can repeat twice can be 0,1;0,2;0,3;0,4;0,5,etc...1,2;1,3;1,4;1,5..etc,2,3;2,4;2,5...etc,2,9; until 8,9 so a total of 45 possibilities.. but out of 10000? because 10^4? Anyways the answer is supposed to be 630.... I don't know how that's possible.. Thanks

2nd problem

From the population of five symbols a, b, c, d, e, a sample of size 25 is taken. Find the probability that the sample will contain five symbols of each kind.

The only thing I could think of was (4/5)^5 x (3/5)^5 x (2/5)^5 x (1/5)^5 which is obviously wrong. Apparently the answer is 25!/(5!^5 X 5^25 ) so if anyone could explain the logic behind this answer would be great! Thank you guys! :)

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# 2 problems in combinatorics

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