From the letters of the word FOREVER, 4-letter codes were formed. How many codes:
(a) were formed?
Case 1: No repeated letters. These are 5 letters (FOREV) PERMUTE 4.
That's 5P4 = 5∙4∙3∙2 = 120
Case 2: With R repeated but not E:
Choose the 2 positions for the R's to go in 4C2 = 6 ways.
That leaves 2 unfilled positions.
Choose the letter to go in the leftmost unfilled position from FOEV
4 ways.
That leaves only 1 unfilled position.
Choose the letter to go in it in 3 ways.
That's 6∙4∙3 = 72
Case 3: With E repeated but not R:
Same as Case 2. 72 ways
Case 4: with 2 R's and 2 E's.
Choose the 2 positions for the R's in 4C2 = 6 ways.
Fill the remaining 2 positions with E's in 1 way.
That's 6∙1 = 6 ways for case 3.
Total from all three cases: 120+72+72+6 = 270 ways.
(b) Begin and end with an R?
These are the arrangements of two letters from FOEVE with an R added
on each end.
Case 1: use both E's between two R's, i.e., REER
That 1 way for case 1
Case 2: use 2 letters from FOEV in between the 2 R's.
That's 
Total for both cases: 1+12 = 13 ways
(c) Had the letters EVER?
If all the letters of EVER were distinguishable it would 4!, but since there
are 2 indistinguishable E's, we divide by 2!, so the answer is
(d) did not have any repeated letters?
That's Case 1 of part (a)
Answer: 120
(e) began and end with a vowel?
Case 1: Both R's between vowels: ORRE, ERRO, ERRE
That's 3 for Case 1.
Case 2: E--E with two different letters between the E's taken
from FORV.
That's
Case 3: E--O with two different letters between the E's taken
from FREV,
That's also 12
Case 4: O--E with two different letters between the E's taken
from FREV,
That's also 12
Total from all four cases: 3+12+12+12 = 39 ways.
Edwin
