Number of $r$ letter words taking letters from a $n$ letter word I can't figure out how to do questions such as this one, any thoughts?

What is the number of four letter words that can be formed from the letters in BUBBLE (each letter occuring at most as many times as it occurs in BUBBLE)?

The answer is $72$, but I am not sure how to arrive there.
 A: Hint: How many four letter words can you make if you include one B? How many four letter words can you make if you include two B's? How many four letter words can you make if you include three B's?
Good luck!
A: To me it would be:
you have to make word of four letter with letters from $\{B,B,B\}$ and $\{U,L,E\}$. So, you take $1$ B and $3$ letter of the 2nd set with $\binom{3}{1}\binom{3}{3}$ possibilities. Then you can make $4!$ words. You can also take $2$ B, then you have to take 2 letter from the other set, and you can make $\frac{4!}{2!}$ words. Finally, if you take $3$ B, you can take only one letter of the other set, and you can make $\frac{4!}{3!}$ words. Finally, you can make
$$\binom{3}{1}\binom{3}{3}4!+\binom{3}{2}\binom{3}{2}\frac{4!}{2!}+\binom{3}{3}\binom{3}{1}\frac{4!}{3!}$$
words, but it makes $192$ possibilities... so I'm probably wrong...
A: You calculate the number of 4-letter words that can be formed if you had 6 different letters. Then you have to find out by how much you "overcounted", i.e. how many words are there where one can let letters change their places without changing the word.
A: *

*Using Combination

In the case where 1 B is present, we can have the word “BULE” which can be arranged in 4! Ways to give all the words.
Where there are 2 B’s, for example “BBUL”, the 2 B’s can be selected along with 2 more letters from {U, L, E} in C(2,2) x C(3,2) ways which can be arrange in 4!/2! to give all the possible words. So there are C(2,2) x C(3,2) x (4!/2!) words.
For words with 3 B’s, selecting the 3 B’s and a letter from {U, L, E} can be done in C(3,3) x C(3,1) ways. So all the possible word form are C(3,3) x C(3,1) x (4!/3!) words.
So there are 4! + C(2,2) x C(3,2) x (4!/2!) + C(3,3) x C(3,1) x (4!/3!) = 72 words


*Using Permutation

The set containing all the letters {B, U, B, B, L, E} can be broken down in two more sets X = {B, B, B}, Y = {U, L, E} where one contains the B’s (similar elements) and the other the contains the remaining letters (distinct elements).
For the case of a B appearing only once in the four letter word, the B can be selected from set X in P(3,1) way and the remaining 3 letters from Y in P(3,3) ways. So there are P(3,1) x P(3,3) words containing 1 B.
For words with 2 B’s, selecting the 2 B’s from X can be done in P(3,2) ways and selected 2 letters from Y in P(3,1) ways. This gives P(3,2) x P(3,2)  words with 2 B’s.
Selecting all 3 B’s and a letter from X and Y respectively can be done  P(3,3) x P(3,1) ways.
So there are  (P(3,1) x P(3,3)) + (P(3,2) x P(3,2)) + (P(3,3) x P(3,1)) = 72
