How many different words can be formed using the letters (m, a, t, h, s)? How many different words of length $5$ can be formed using all the letters m, a, t, h, s without repetition 
If the first letter must be a vowel?
If the first letter must be a consonant?
It is a question from my book in the permutations chapter. I have no idea how to approach it. I know about factorials but I get confused when they specify the vowel and the consonant cases.
 A: If the first letter must be a vowel: you have the "a" as first letter and you can permutate 4 letters, thus you can create $4!$ words (meaningless I suppose)
If the first letter must be a consonant: you have 4 possibilities for the first letter and you can permutate the remaining 4 letters, thus you can create $4\cdot4!$ words
NOTE: as a check summing up the two results you obtain $$4!+4 \cdot 4!=5 \cdot 4!=5!$$ that is the total number of permutation for the 5 letters.
A: I assume your book means permutations, not English words.  If it means English words the question has a lot less to do with maths.  To find the answer, notice that there are five letters in maths, so there are five different letters that can go in the first position, either m can come first, or a, or t, or h, or s.  Next, you've used one of these letters, so only 4 are available to come second in the word.  Now you've used 2 letters, so only 3 are available to go third.  This continues, so the answer is 5x4x3x2x1=120.  Now if you want to find the number of ways you can do it with consonant first, your question is more complicated.  For vowels, it is easy, as there is only one letter that can come first: a.  A is in spot one.  There are four options for spot 2.  3 options for spot 3.  2 for 4, and 1 for 5.  The answer is 4x3x2x1=24.  For starting with a consonant, 4 letters can come first, because a can't, and now you only have 4 letters left, so only 4 letters for spot 2, 3 for 3, 2 for 4, and 1 for 5.  The answer then is 4x4x3x2x1=96.
