I am faced with the following problem:
a) Find the amount of permutations which can be obtained from the letters of the word $MUSICA$
b) How many of these permutations begin with a vowel?
c) How many begin or end with a vowel?
d) How many of them have all of the letters in a different place than the original word?
e) How many include the word $MAS$?
For a) and b) I quickly came to the conclusion that the answers are $6!$ and $3 \cdot 5!$ respectively. My explanation for this second result is that there are three vowels fixed in the first position, and $5!$ permutations to be made from the remaining letters.
In c) I'm already having a harder time... I would conjecture that the amount of permutations that begin and end with a vowel are both going to be $3 \cdot 5!$ and that I must then subtract those that begin and end with a vowel. I'm having a hard time visualizing this last quantity though.
d) has me a bit stumped. I think I have to go about it a backwards way, like counting the words that are in the right place and subtracting them from the total, but I'm not sure how to count that!
I know from my teacher's tip that I need to apply the Principle of Inclusion and Exclusion, but I am having a hard time of it. I know what the Principle is but I do not see how to directly apply it here, even though I kind of see how it applies to what I conjectured as an answer for c).
Finally in e) my idea is to count MAS as if it were a letter and thus there must be $4!$ permutations of $MUSICA$ which contain $MAS$.
How am I doing? Thanks SE.
You're doing fine!
a and b correct!
For c: correct approach. Number that you need to subtract is $3 \cdot 2 \cdot 4!$ ($3$ choices for first letter a vowel, which leaves two choices for last letter a vowel, and of course $4!$ permutations for the rest)
For d: look up derangements. The number of derangements for $6$ letters is $!6=265$ ... to get that number using the Inclusion-Exclusion Principle: take all possible arrangements ($6!$) but subtract all the ones that have the first letter in the first position ($5!$), subtract all the ones that have the second letter in the second position (also $5!$), and of course do this for all $6$ letter positions, i.e. We get $6! - 6 \cdot 5!$. OK, but now we need to add back in those that have both the first and the second letter in the right position (of which there are $4!$), since we subtracted those twice, and we do this for all $6 \choose 2$ pairs, so we get $6! - 6 \cdot 5! + {6 \choose 2} \cdot 4!$. Ok, but now we added too many again (e.g. We added the ones with the first, second, and third letter all in the right position several times), so now subtract $3!$ permutations for each of the $6 \choose 3$ words with three letters in the right position. .... and you see how this continues ... In sum, we get:
$$!6=6! -6 \cdot 5! + {6 \choose 2} \cdot 4! - {6 \choose 3} \cdot 3! + {6 \choose 4} \cdot 2! - {6 \choose 5} \cdot 1 + {6 \choose 6}=$$
$$720-6 \cdot 120+ 15 \cdot 24- 20 \cdot 6 + 15 \cdot 2 - 6 \cdot 1 +1=$$
$$720-720+360-120+30-6+1=265$$
e) all good!