Question 1207461
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Please resolve this question. You have two 10 cents coins, two $1 coins, one fifty cents coin and a $2 coin. 
one coin is randomly chosen and is not replaced. two more coins are then randomly chosen, 
What is the total number of possible outcomes for this chance experiment.
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        The solution by Edwin is incorrect.
        It is incorrect, since Edwin counts many identical outcomes as different outcomes.


        I apply totally different logic and obtain totally different solution and answer.



<pre>
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 {{{C[2+2+1+1]^3}}} = {{{C[6]^3}}} = {{{(6*5*4)/(1*2*3)}}} = 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.    <U>ANSWER</U>
</pre>

Solved.


Hope Edwin will agree with this solution,
and we will celebrate the triumph of reason together.