Let's look at a different question first: how many ways are there to put $25$ bars between $10$ stars? (when between every two stars there can be as many bars as I wish). You need to take a line of $35$ places, choose the $10$ places where you want the stars to be, in the rest of the places there will be bars. So you just need to choose $10$ places from $35$ which is $\binom{35}{10}$ options.
Now, why is it equivalent to your problem? Because to create a word you need to choose how many times the letter $a$ will appear (zero times is also an option), how many times $b$ will appear and so on. Once you choose that for every letter you will get the word because you are not allowed to choose the order of the letters. So the number of times $a$ will appear is the number of stars before the first bar, the number of times $b$ will appear is the number of stars between the first and second bars and so on. The letter $z$ is letter number $26$ so it is the number of stars after the $25$th bar. So the number of words is $\binom{35}{10}$.
There are 47 sets of 4 consecutive elements, from $\{1,2,3,4\}$ to $\{47,48,49,50\}.$ For each of these, we can choose a fifth element for our 5-tuple in 46 ways, and this fifth element can be placed in the first or last position in the 5-tuple. So far our count is
$$47 \cdot 46 \cdot 2$$
Notice this count would include the 5-tuples with five consecutive elements as well; however each of these 5-tuples with 5 consecutive elements would have been counted twice, either by choosing as our fifth element the first or the last of the five consecutive numbers. [e.g the 5-tuple $(23,24,25,26,27)$ could be created by placing $23$ in front of the 4-tuple $(24,25,26,27)$ or by placing $27$ at the end of the 4-tuple $(23,24,25,26)$.]
Thus we need to subtract the number of 5-tuples with 5 consecutive elements, of which there are clearly 46. So our final count will be
$$47 \cdot 46 \cdot 2 - 46 = \boxed{4278}$$
N.B. I've assumed in the above that the 5-tuples are to be made from five different elements; if repetition were allowed, then in the first part of the computation, when choosing the fifth element there would be 50 choices rather than 46, so then the count would be
$$47 \cdot 50 \cdot 2 - 46 = \boxed{4654}$$
Best Answer
Let $a_1$ be the number of $1$'s that we use, and $a_2$ the number of $2$'s, and so on. Then our $5$-tuple is completely determined once we know $(a_1,a_2,\dots,a_n)$. So we want to count the number of solutions of $a_1+a_2+\cdots+a_n=5$ in non-negative integers.
By Stars and Bars (please see Wikipedia if the term is unfamiliar) the number of solutions is $\dbinom{5+n-1}{n-1}$ or equivalently $\dbinom{n+4}{5}$.