Even Number cards? There are $15$ cards on a table, marked with an integer $1$ from to $15$ . How many ways can I take cards such that the sum of the numbers on the cards is even?
Please help me?
 A: Generating function approach.
Let $$f(x)=\prod_{i=1}^{15} \left(1+x^i\right)=\sum_{j=0}^{15\cdot 16/2} a_jx^j$$
where $a_j$ is the number of ways of getting a total of $j$ from taking cards.
Then you are looking for $\frac{f(1)+f(-1)}{2}$. But $f(-1)=0$, so you are looking for $\frac{f(1)}{2}=2^{14}$.
That includes taking zero cards to get zero, so you might want the answer $2^{14}-1$.
A: Using the fact that sum of two odd cards is even or a single even card is even.If you want even cards you must take these in these quanta.
The number of ways in which you can take even cards(total 7) is [none or more]:
$$2^7=128\text{ ways }$$
The number of ways in which you can take pairs of odd cards(total 8 cards, 4 pairs at a time) is same as taking even quantity of odd faced cards, i.e.:
$$\sum_{k=0}^{4}\binom8{2k}=2^{7}=128\text{ ways }\quad\left(\because \sum_{k=0}^{n/2}\binom n{2k}=2^{n-1}\right)$$
Total $2^7.2^7-1=2^{14}-1=16383$, [ minus one the case in which both quanta have zero cards.]
Similiarly it can be showed that for n cards it is $2^{n-1}-1$
A: Hint: 7 cards have even numbers and 8 cards have odd numbers.  You can take any  of the even cards, and you must take an even number of the odd cards.  Calculate these two amounts separately, then multiply.
A: Hint: For any subset of the first $14$ cards ($1, 2,..., 14$), there is exactly one way to complete it by using or not using card $15$, so as to make the sum even.
A: An opportunistically easy Olympiad-style smartass proof of the general result derived by Aditya (which we could also have gotten directly from symmetry of Pascal's triangle)


*

*In the general case with n cards, there are $2^n$ ways of picking a subset partition (for now, include the empty set).

*In our specific case n=15, there is an even number of odd cards, of which we are constrained to pick an even-numbered subset of odd cards, and any number of even cards.

*But note that for every valid partition $P = \{p_i\}$ there exists exactly one corresponding partition $P'$ which is invalid; we can choose to construct $P'$ by either $P' = U \setminus P$ or $P' = \{15 - p_i\}$. (Both are invalid because both change the number of odd cards (since n=15 had an odd number of odd cards. If n was even this would be slightly harder).

*Hence $#P = #P'$. Since $#P + #P' = 2^n$, it follows that $#P = 2^n-1 -1$ (where we finally exclude the empty set).

