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i.) How many subsets of $A$ contain six elements?

ii.) How many six-element subsets of $A$ contain four even integers and two odd integers?

iii.) How many subsets of $A$ contain only odd integers?


I need hints. I'm stucked on the first one, maybe after I do it I can make progress with the other, but feel free to leave hints for the others too. Thanks.

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    $\begingroup$ Hint: think combinations: you're just finding how many ways to pick some things from a group of things $\endgroup$ – Brent Mar 7 '15 at 2:21
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For i.): Choosing a certain number $m$ of elements out of a set of $n$ without regard for the order is calculated by $n \choose m$. In this case it is $12 \choose 6 $$=924$.

For ii.): Possibilities to get 4 even numbers out of 6 is $6 \choose 4$$=15$ and there are $6 \choose 2$$=15$ ways to get 2 odd numbers out of 6 (which is by the way the same number because you can reformulate it to not count the ways to draw 4 but to throw out 2). To get the final number of ways we have to multiplicate both, so we get $15\cdot 15=225$.

For iii.): There are 6 odd integers in the set, you decide for each odd number if you include it in a set or not, that is 2 choices - and as there are 6 odd numbers you get to choose 6 times. You can get each of the described sets with this method and you miss none if you play through all choices. So the total number of sets containing only those is $2^6=64$.

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    $\begingroup$ Aren't there only $6$ odd integers in $A$ -- $1,3,5,7,11,17$? Maybe I miscounted. $\endgroup$ – kobe Mar 7 '15 at 2:47
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    $\begingroup$ Does the empty set contain odd integers? $\endgroup$ – N. F. Taussig Mar 7 '15 at 2:48
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    $\begingroup$ @Paul thanks for your answer, but I think I got it now. It must be $2^6-1$ since there are $2^6$ ways to list the odd integers. But I gots to subtract the empty set from $2^6$ so $2^6-1$ cheers. $\endgroup$ – user265675 Mar 7 '15 at 2:48
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    $\begingroup$ @N.F.Taussig: Does the empty set contain only odd integers? (Yes.) $\endgroup$ – Nick Matteo Mar 7 '15 at 2:54
  • $\begingroup$ @cobe: I miscounted... shame on me, it's corrected now $\endgroup$ – Paul Mar 7 '15 at 2:57

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