Why is this a combination problem, when order clearly matters? I have an exam in discrete mathematics tomorrow and I am really having a hard time wrapping my head around some of the combination and permutation problems. For instance this question underneath, I have absolutely no idea how to solve! In my opinion it should be a permutation problem, as order matters, but somehow non of the solutions involve permutations. Why?
(g) (2 pt.) An ITU username is lucky, if it has exactly four characters (there are 26 characters from a-z), and three consecutive characters are identical. E.g.,


*

*aaab is lucky, since there are three consecutive a’s.

*ccbc is not lucky; even though there are three c’s, they are not consecutive.

*beee is lucky, since there are three consecutive e’s.

*abdc is not lucky
How many lucky usernames are there?
(1) 67600 = (26!/3!(26-3)!)*26
(2) 5200 = (26!/3!(26-3)!)*2
(3) 1300 = (26!/2!(26-2)!)*4
(4) 650 = (26!/2!(26-2)!)*2
Thanks!
 A: All possible answers seem to be based on selection.  So...
You seek the ways to select two from twenty-six letters, select which one from the two to be the single, and select one from the two ends to place that single.
Which answer corresponds best to that?
A: Note that as three consecutive letters should be identical, we can select this identical element in $26$ ways. The non-identical element can be then selected in $25$ ways. 
But, there are two ways to place this non-identical element, either in the first place or at the last, giving us a total of: $$26\times 25 \times 2 =1300$$
A: Only lucky usernames with $a$ are: $$aaa\chi\quad \chi aaa$$
where $\chi$ can be $any$ one of the remaining 25 letters, i.e. number of lucky usernames with $a$ are $2\cdot 25$. This is true for every initial choice from one of the 26 letters, not just $a$, and thus $$2\cdot 25\cdot 26$$
is the number of lucky usernames.
A: It is true that order matters here, and this is a "permutation" rather than "combination" question. You should start off by choosing the letter to appear three times and the letter to appear once (in order). There are ${}^{26}\mathrm P_2$ ways to do this; however, ${}^{26}\mathrm P_2$ could also be written as ${}^{26}\mathrm C_2\times 2$. Then you need to multiply by the number of lucky usernames with three of the first letter and one of the second (i.e. by $2$, since aaab and baaa are the possible lucky names with three as and a b).
The question writers have put the answers in a strange form to make it harder to tell them apart; part of the problem is spotting which one of those is equivalent to the form you would naturally get it in.
