consider strings of length 10 which contain only letters from the set{A,E,I,O,U} and digits from {1,3,5,7,9}suppose repetition of letters is not allowed.
a) how many different strings are there?
P(10,10) = 10! = 3628800
b) How many different strings are there if the letters,i.e A,E,I,O,U and the digits,i.e 1,3,5,7,9 must alternate?
They can go either
LDLDLDLDLD or DLDLDLDLDL, where D stands for a digit and L for a letter.
There are P(5,5) = 5! = 120 ways the letters can be arranged times
P(5,5) = 5! = 120 ways the digits can be arranged. Then we multiply by
2 because we can either start with a letter or a digit.
Answer: 5!*5!*2 = 120*120*2 = 28800.
c) How many different strings are there if all five letters must be adjacent in each string?
Think of (AEIOU) as a single thing. Then there are these 6 things
to rearrange:
1,3,5,7,9,(AEIOU)
which is P(6,6) = 6! = 720. But we must multiply this by P(5,5) = 5! since the
letters (AEIOU) can be rearranged in 5! ways.
Answer 6!*5! = 720*120 = 86400
Edwin