How many $5$-letter words have $2$ or more C's in them If I want to calculate how many ways there are to create a string 5 letters (26 to choose from) long that contains at least 2 C's, would I do ${5 \choose 2}\cdot 26^3=175760$? That picks spots for at least $2$ C's, then the rest can be any letter? Why is this not the same as adding up each case (strictly 2 C's + strictly 3 C's +...)? That would be $\sum_{i=2}^5({5 \choose i}\cdot 25^{5-i})=162626$. Why the difference and which is correct?
Thanks for the help.
 A: Your second method is correct.  Here is another method:
There are $26^5$ five-letter words.  From these, we subtract those with fewer than two Cs.  There are $25^5$ words with no Cs and $\binom{5}{1}25^4$ words with exactly once C.  Hence, there are
$$26^5 - 25^5 - \binom{5}{1}25^4$$
five-letter words with at least two Cs.
As @peterwhy indicated in the comments, your first method counts each word with more than two Cs multiple times.  You count each word with three Cs three times, once for each of the $\binom{3}{2}$ ways you could designate two of the three Cs as the two Cs in the word; each word with four Cs six times, once for each of the $\binom{4}{2}$ ways you could designate two of the four Cs as the two Cs in the word; and the word CCCCC ten times, once for each of the $\binom{5}{2}$ ways that you could designate two of the five Cs as the two Cs in the word.  Observe that
$$\binom{5}{2}25^3 + \color{red}{\binom{3}{2}}\binom{5}{3}25^2 + \color{red}{\binom{4}{2}}\binom{5}{4}25 + \color{red}{\binom{5}{2}}\binom{5}{5} = \color{red}{\binom{5}{2}26^3}$$
The source of the error in your first method is that the two Cs and the three additional letters are not disjoint since there may be Cs among the additional letters.  In your second method, you will notice that your additional letters are disjoint from the Cs you select.
A: As N. F. Taussig's answer says, the second method is correct.
Here's an intuitive (hopefully) way to think of why the first method is incorrect:
The first factor $5 \choose 2$ counts the number of positions that $2$ C's can take in the $5$ letter word. Think of an "incomplete" word where only the $2$ $C$'s are placed: $CXXCX,XCXXC,CCXXX,...$
The second factor $26^3$ is counting how many possible "complete" words can be constructed for each of the "incomplete" words by choosing out of the $26$ available letters. e.g $CABCD,VCGAC,CCNPH...$
If we happen to choose the letter $C$ when constructing the "complete" words from the "incomplete" ones, then there is a chance of overcounting. For example, the words $CACCB$ and $CACCB$ could technically be different since their original "incomplete" forms could be either $CXXCX$ or $CXCXX$.
Hope this helps :)
