Find the number of increasing words of length $n$ formed by an alphabet of $m$ letters 
Prove that the number of increasing words of length $n$ formed by an alphabet of $m$ letters is $$\binom{m+n-1}{n}$$
(A word is increasing if its letters(except repetitions) appear in alphabetic order)
$e.g.: abbbbccdeeef$ is an increasing word of length 12

I would really appreciate your help; I don't know how to solve this :(
 A: It's all about finding the right bijection into a set you can count.
Starting with an alphabetical string, form an auxiliary string as follows. Between consecutive blocks, place a barrier symbol $|$. Replace each letter by a $_$. For instance, you string $abbbbccdeeef$ becomes
$$
_|____|__|_|___|_
$$
Do you see why there is a bijection between these auxiliary strings and the alphabetized ones we started with?
Now we count the auxiliary strings. In total, there are $m-1$ barriers and $n$ spaces $_$. Hence we've written down $n+m-1$ symbols. Now we choose the locations for the $m-1$ barriers. Hence there are
$$
{n+m-1 \choose m-1}
$$
possible auxiliary strings. Now since $n=(n+m-1)-(m-1)$, it follows that
$$
{n+m-1 \choose m-1}={n+m-1 \choose n}
$$
which is the answer we desired.
A: Step 1: make $n$ choices from an $m$-letter alphabet, with repetition allowed and order not important.  This a standard "dots and lines" ("stars and bars") problem and the number of selections is
$$\binom{m+n-1}{n}\ .$$
Step 2: arrange your letters in alphabetical order.  There is only one way to do this!!  So the answer is the number we have already.
A: Notice that we can encode the word $abbbbccdeeef$ by using a string of stars and bars as follows:
$$
\star | \star \star \star \star | \star \star | \star | \star \star \star | \star
$$
where the $m - 1$ bars separate the string into $m$ slots (one for each letter in the sorted alphabet) and the $n$ stars represent the total number of letters. Hence, counting the number of desired words is tantamount to counting the number of possible strings we can make using $n$ stars and $m - 1$ bars. This yields:
$$
\frac{(n + m - 1)!}{n!(m-1)!} = \binom{n + m - 1}{n}
$$
as desired.
