Originally Posted by

**Plato** I confess that I am not convinced that I read this correctly.

For example I assume that saying "overlapped elements" means that elements can be repeated. In that case the answer is indeed $3^3=27$

Now for the second part, this is a pure guess. "how ways can one select three items from three types where repetitions are allowed?"

For this part, part 2: $<1,2,2>~\&~<2,1,2>$ are **not different**, where as $<1,2,2>~\&~<2,1,2>$ **are different** in part 1.

Think of three labeled boxes $~\boxed{~1~},~\boxed{~2~},~\boxed{~3~}$ along with three identical balls, $o,~o,~o$.

How many ways can we place the halls into the boxes? Here are three ways:

i) $o~|o~|o$ one in each box;

ii)$~|oo~|o$ none in box 1, two in box 2, & one in box 3;

iii) $ooo~|~|$ all in box 1.

Now if we simply count the number of ways to arrange the string $~|~|~o~o~o~$: $\dfrac{(2+3)!}{(2!)(3!)}=10$

In general, if we have $N$ different types of objects (each type having $K$ is stock)

the number of ways to select a collect of $J\le K$ of these objects is $\dfrac{(N+J)!}{(N!)(J!)}$

This idea can answer a wide variety of questions.

1) How many non-negative integers solve $s+t+u+v+w+x+y+z=100~?$

2) Using the letters $\{A,B,C,D,E,F,G,H,I,J,K\}$ how many strings of five letters, allowing repetitions, are in alphabetical order? etc.