consider the collection of all strings of length 10 made up from the alphabet 0,1,2 and 3. how many have even weight? 
Consider the collection of all strings of length 10 made up from the alphabet 0,1,2 and 3. how many have even weight?

This is very weird question for me. According to solution the answer is $$
2^{10}\cdot\sum_{i=0}^5 \binom{10}{2i}
$$ which isn't making sense to me. According to me it is supposed to be done in cases like all 0's Or 2's (even). Two odd rest even four odd rest even and so on definitely it will be extremely large and there must be another way around for which I am here. But how is the upper solution even possible as length of string is 10 and if i will be 5 then do they mean as 0's and 2's are represented by $2^{10}$.
This $2^{10}$ multiplied by that combination of (10,10) ? But why. How's this possible. 
 A: First, you have six possibilities: no odd numbers in the string, two odd numbers in the string, four odds, six odds, eight odds or ten odds. For each of those possibilities, you will have to choose where in this string of $10$ you want to place those odd numbers. If you have no odd numbers, you can do that in $\binom{10}0$ ways. If you have two odds, you can choose where to place them in $\binom{10}2$ different ways. If you have four odds, you can...
All in all, the total number of ways to choose where to place the odd numbers is
$$
\binom{10}0 + \binom{10}2 + \binom{10}4 + \binom{10}6 + \binom{10}8 + \binom{10}{10} = \sum_{i = 0}^{5}\binom{10}{2i}
$$
Hopefully this looks familiar. It turns out to be equal to $2^9$ (you can freely choose even or odd for each of the nine first spots, and then the final spot is forced), but that's apparently not exploited in this case.
Now that we have decided where to place the odd numbers (and, by extension, the even numbers), we just need to pick, for each of the ten spots, one of the two possible numbers of the appropriate type. This can be done in $2^{10}$ ways, no matter which of the above $\sum_{i = 0}^{5}\binom{10}{2i}$ even-odd distributions we choose. So in total, we have
$$
2^{10}\times \sum_{i = 0}^{5}\binom{10}{2i}
$$
different ways of doing this.
There is an easier way to count this, however. And that is to note that exactly half of all the possible strings have even weight. In fact, if you take a string with even weight, and you swap the last digit for $3$ minus that digit (so a $0$ turns into a $3$, a $1$ turns into a $2$, and so on), then you get a string with odd weight. And if you carry out the same operation again, you get the original string back. This way, you can "pair up" each even weight string with a corresponding odd weight string (that only differs in the final digit), and we see that the number of even weight strings is equal to the number of pairs. And the number of pairs is $\frac{4^{10}}2 = 2^{19}$.
Or, even easier: Place the first nine digits in the string. That can be done completely freely, without any restrictions, so there are $4^9$ ways. Then whether the last one is even or odd is forced, and you only have $2$ options, giving a total of $4^9 \cdot 2$ possibilities.
