How many arrangements of the word BANANAS are there where the $3$ A's are separated?
I know that once chosen the places for the three A's, there are $\dfrac{4!}{2!}=12$ possible arrangements for the rest of the letters (we divide by $2!$ because there are $2$ N's). But I am having trouble with choosing the places for the A's.
If I do this manually, I count $10$ different arrangements for the $3$ A's, and that would mean that there is a total of $12\cdot 10$ possible arrangements that fit the initial condition. However, I would like to learn to calculate the $10$ cases with a combinatorics argument, instead of just counting. Could someone help me?
from Hot Weekly Questions - Mathematics Stack Exchange
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