The values of the coins are irrelevant. Also, the probabilities of
selecting coins is also irrelevant.  [In future problems like this
they will be relevant].  

But all that matters is that you started with 6 coins, two pairs 
and two singles, where each member of each pair is indistinguishable 
from the other member of that pair.  Say, the 6 coins are AABBCD.

Case 1. You choose one of the A's not to replace. You have left ABBCD.
You can then choose AB,AC,AD,BB,BC,BD,CD
There are 7 outcomes.  

Case 2. You choose one of the B's not to replaced.  Same as case 1. 
There are 7 outcomes.

Case 3. You choose the C not to replace. You have left AABBD.
You can then choose AA,AB,AD,BB,BD
There are 5 outcomes.

Case 4. You choose the D not to replace.  Same as case 3. 
There are 5 outcomes.

Total = 7+7+5+5=22 possible outcomes.

Edwin

The problem's question is the same (= is equivalent) as to ask 


    having two 10-cent coins, two 1-dollar coins, one 50-cent coin and one 2-dollar coin,
    in how many different ways can you choose a triple of coins (without looking on the order).



Had all 6 coins be different, we would choose  =  =  = 5*4 = 20 
different triples.


But some triples are indistinguishable, because they have indistinguishable coins;
therefore, we should count indistinguishable triples only ONCE, NOT twice.


So, you have  the set  AABBCD  to start choosing triples from it.


All unique distinct triples are

    ABC    (no repetition)

    ABD    (no repetition)

    ACD    (no repetition)

    BCD    (no repetition)


    AAB    (with repeating A)

    AAC    (with repeating A)

    AAD    (with repeating A)


    ABB    (with repeating B)

    BBC    (with repeating B)

    BBD    (with repeating B)


Thus these 10 outcomes are all possible different outcomes in this experiment.    ANSWER

It depends on what you mean by "outcome"   

I took "outcome" to mean "situation".
Ikleyn took "outcome to mean "coins drawn"

For instance, I considered A|BB and B|AB as two different outcomes.
She took them as the same outcome.

I guess she's right.  I possibly misinterpreted what "outcome" meant.

Edwin