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December 17th, 2015, 07:36 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85  How Many Combinations Of The Numbers From 1 To 6?
Six numbers from 1 to 6 are selected with every number equally likely to be selected on every independent trial. Any number can be selected any amount of times from 0 to 6. 1. How many combinations of numbers are possible? 2. How many combinations of numbers are possible if frequencies matter but not what number? For example, selecting the same number all six times will count as 6 combination in Question 1 but as 1 combination in Question 2. 
February 26th, 2016, 04:23 PM  #2 
Member Joined: Apr 2014 From: Birmingham, AL Posts: 36 Thanks: 4 
Great Question! 1. How many combination of numbers are possible? So at first I was like oh, easy, it's (6^6)/6! But I was wrong.. So I'm pretty sure the answer is n+r1 C r1 or 11 C 5 = 462. I found two sources to help me. The trouble is, they don't agree with what the solution should be! Well, actually for this particular n and r, they agree. But that's just coincidence. This intuitive answer suggests that the answer to this question should be n+r1 C r1. This makes sense to me. However, on wikipedia, I'm almost positive that they say the answer to the same question should be n+r1 C r. For n=6, r=6, this happens to be the same as the first source. However, that's just a coincidence. Would love if someone could explain to me what I am missing here.. 2. You misworded question 1. The word combination means that order doesn't matter i.e. {1,1,1,1,1,2}={2,1,1,1,1,1}. So what you were looking for with this one was 462 or 11 C 5. What you were looking for in the first one was (n+r1)!/(r1)! = 332640  same as combinations except now we want to count each "1" that's selected as different so we do not divide by n!. 
February 28th, 2016, 07:13 AM  #3 
Member Joined: Apr 2014 From: Birmingham, AL Posts: 36 Thanks: 4 
For anyone interested in the difference between those two formulas, the difference is that in wikipedia source you are distributing k items into n bins. So the formula is k+n1 C n1 = k+n1 C k+n1(n1) = k+n1 C k. In the other source, we are distributing n items into k bins. In that case the formula is n+k1 C k1. 

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