How many 4-letter words can I make with the letters $\{W, X, Y, Z\}$ if the letter $W$ must be used an odd number of times? Assume I can repeat letters.
How do I create cases to count W for odd number of times it is used? Is there another way to solve this without using casework?
On
Three letters $X,Y, Z$ can be placed by $3\cdot 3\cdot 3$ ways. The letter $W$ can be placed at any of four available places by $4$ different methods. Therefore, total number of words having four letters out of $\{W, X, Y, Z\}$ such that $W$ comes once only, will be $$3\times3\times3\times4=108$$
Similarly, three $W$'s can placed at three places by $\dfrac{4!}{3!}$ ways and empty fourth place can be filled by any of $X, Y, Z $ letters by $3$ ways. Therefore total number of words having four letters out of $\{W, X, Y, Z\}$ such that $W$ comes $3$ times, will be $$=\frac{4!}{3!}\times 3=12$$ Hence the total number of words required $$108+12=120$$
Let's solve this problem for the $n$-letters words.
Let's call $A_n$ the $n$-letter words with an odd number of $W$ and let's call $B_n$ the $n$-letter words with an even number of $W$. We know that $$ A_n = 3A_{n-1}+B_{n-1}$$ and that $$B_n=3B_{n-1}+A_{n-1}$$ (think to the last letter you add).
Moreover, you know that $A_1=1$, $A_2=6$, $B_1=3$, $B_2=10$.
Knowing this you can create a table and fill it using this rules.
(EDIT: I think that working on these two equation MAYBE you can create a closed formula but idk)