Suppose I pick two number from {0,1,2} without replacement and suppose I keep track of which one was drawn first. This is a permutation question.(adsbygoogle = window.adsbygoogle || []).push({});

There are 3!/(3-2)! = 6 possible permutations:

(0,1)

(0,2)

(1,0)

(1,2)

(2,0)

(2,1)

Of course, if I only care which numbers were choosen, then I care about combinations.

THere are 3!/(3-2)!/2! = 3 possible combinations.

(0,1)

(0,2)

(1,2)

All very basic....

Now, I ask the same question with replacement. Now, the drawings are independent. 3^2 = 9

(0,0)

(0,1)

(0,2)

(1,0)

(1,1)

(1,2)

(2,0)

(2,1)

(2,2)

Suppose I don't care about permutations. That is to say, I just want to know what numbers were picked with replacement.

(0,0)

(0,1)

(0,2)

(1,1)

(1,2)

(2,2)

Now there are only six combinations possible. How do I get this number? The same thing in terms of head/tails:

00

01

10

11

4 possibilities when we care about order, but only 3 when we do not since 01 is 10. I know this must be very simple, but I don't know the formula. I am asking: If I flip a coin twice, what is the number of "unique" outcomes. It is easy with a coin....the number is always associated with the total number of possible heads (or tails)...which is always 0 to n. But how do I do this in general if there are more choices.

If I draw k items from a set of n items with replacement, then how many different combinations (not permutations) are possible?

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