How many words can be formed by taking 4 letters at a time from the letters of the word 'MORADABAD'?

Question

What I tried was: (9P4)/3!*2! This gave me a wrong answer (since the answer is 626).

I'm unable to make use of the hint provided in my book: "make cases".

Any help would be appreciated. :)

Answer 1

As the hint suggests, divide into cases, and count carefully.

Case 1: All $4$ letters distinct. There are $6$ different letters in our long word. We can arrange $4$ of them in a row in $(6)(5)(4)(3)$ ways. Alternately, e can choose the $4$ distinct letters in $\binom{6}{4}$ ways, and for each way arrange them in $4!$ ways, for a total of $\binom{6}{4}4!$.

Case 2: Two A's, the rest distinct. We can choose the places where the A's go in $\binom{4}{2}$ ways. For each such choice, there are $(5)(4)$ ways to fill the remaining slots with two distinct letters chosen from the remaining $5$.

Case 2': Two D's, the rest distinct. Same analysis, answer as Case 2.

Case 3: Two A's, two D's. The places where the A's go can be chosen in $\binom{4}{2}$ ways.

Case 4: Three A's. We can choose where the A's go in $\binom{4}{3}$ ways, and the remaining slot can be filled in $5$ ways.

Add up.