Arranging a word This is the question :
In how many ways you can arrange the word AAABBCDEFG so that the first letter is A or E ?
I'm not sure if im doing this right. My plan is to take all the arrangments and reduce < br>
the number of options where A is the first letter and than reducing the options where 
E is the first letter and thus getting :
$$\frac{10!}{3!2!} -  \frac{7\cdot 9!}{3!2!} - \frac{9!}{3!2!}$$
sorry about the formating, still new at this, hope it's clear.
Thanks in advance
 A: Here's another possibility:
Choose $6$ positions for BBCDFG, but excluding the first position.  ($9\choose 6$ ways.)
Arrange BBCDFG in these $6$ chosen positions: ($\frac{6!}{2}$ ways.)
Arrange AAAE in three remaining positions ($4$ ways.)
Total: $84\cdot 360\cdot 4=120960$ arrangements.
A: The number of ways to arrange it so that the first letter is A:
Count the number of different permutations of AABBCDEFG:
$$\binom92\cdot\binom72\cdot\binom51\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=90720$$

The number of ways to arrange it so that the first letter is E:
Count the number of different permutations of AAABBCDFG:
$$\binom93\cdot\binom62\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=30240$$

Hence the total number of ways to arrange it so that the first letter is A or E:
Add the number of different permutations of AABBCDEFG and AAABBCDFG:
$$90720+30240=120960$$
A: The ways of arranging AAABBCDEFG without restriction I think you correctly identify as $\frac{10!}{3!2!}$ because of the repeated As (3) and Bs (2).
Fixing an A first gives you $\frac{9!}{2!2!}$ arrangements, and fixing an E first gives $\frac{9!}{3!2!}$ more arrangements. Total is therefore $$ \frac{9!(3+1)}{3!2!} =  \frac{4\cdot 9!}{3!2!} = \frac{9!}{3} = 120960$$
