In the word we have: 1S 1P 1R 2T 2Z 4A
For the first "ending with A" Let's give it a fixed position so let's calculate the anagrams of SPRTTZZAAA
$$\dfrac {10!}{2!2!3!} = 151 200$$
Similarly for starting and ending with $``A"$
$$\dfrac {9!}{2!2!2!} = 45 360$$
I took this exercise form a book which has also it's solutions. But I really do not agree with their solution to this problem because it's contradictory. What do you think of mine? is it valid?
In my manual:
They find $\frac {10!}{3!2!2!}=151200$ Which are the anagram with ending "A"
Total anagrams $\frac{11!}{4!2!2!}=415800$
And say that the anagrams ending with "A" are $415800-151200=264600$ (First problem)
Then calculate the anagrams not ending with A which are $415800 - 151200 = 264600$
And finally they define the ensemble S and E for Starting with "A" and Ending with "A"
Then $S \cap E$ is what I got, $45360$ But they calculate after $S\cup E = |S|+|E|-|S\cap E|= 257040$
Therefore the one starting and ending with "A" are $|\Omega|-|S\cup E|$ (Second Problem)
Is therefore the correction wrong?