Question 1139410
From the letters of the word FOREVER, 4-letter codes were formed. How many codes: 
(a) were formed?
<pre>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. 
</pre>(b) Begin and end with an R?<pre>
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 {{{4P2 = 4*3 = 12}}}

Total for both cases: 1+12 = 13 ways

</pre>(c) Had the letters EVER?<pre>

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
{{{4!/2! = 24/2 = 12}}}

</pre>(d) did not have any repeated letters?<pre>
That's Case 1 of part (a) 
Answer: 120</pre>
(e) began and end with a vowel?<pre>
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 {{{4P2 = 4*3 = 12}}}

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</pre>