If we consider the word $ABCDEFGGGG$. To find the number of arrangments for that word, we just calculate: $\frac{10!}{4!}$.
But if now we want to find the total number of arrangements for that word such that $2$ $G$'s must come together and the two other $G$'s be separated. One of the arrangements is for example: $ABGGCDGEFG$.
*Note that the two G's that are separated must also be separated from the other 2'G that are together.
How can we think about this problem?
Any help will be very appreciated.
Best Answer
The $G$'s are the annoying part. Do the easy part first! There are $6!$ ways to arrange the letters that are not $G$.
There are then $7$ 'spaces', if we count spots to the extreme left or right of the formed word, for other things. Pick three of them. That gives ${7 \choose 3}$ options. One of them will get the double $G$'s, and there are $3$ choices for this. The other two will get individual $G$'s.
This gives a final tally of $(3)6!{7\choose 3}=75600$ ways to meet the stated requirements.
I counted them all. It's my thing, after all.