SOLUTION: Find the number of ways of arranging all 12 letters of word STRAWBERRIES where the first and the last letters are vowels.

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Question 1175422: Find the number of ways of arranging all 12 letters of word STRAWBERRIES where
the first and the last letters are vowels.
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The vowels in the word are A, E, E, and I.

You need to do the calculations as a series of different cases, depending on which vowels are the first and last letters.

(1) The first and last letters are A and I.

Those two can be in either of two orders; the remaining 10 letters include 3 R's, 2 S's, and 2 E's. The number of arrangements for this case is

%282%29%2810%21%2F%28%283%21%29%282%21%29%282%21%29%29%29

(2) The first and last letters are A and E.

Those two can be in either of two orders; the remaining 10 letters include 3 R's and 2 S's. The number of arrangements for this case is

%282%29%2810%21%2F%28%283%21%29%282%21%29%29%29

(3) The first and last letters are I and E.

This case will have the same number of arrangements as case (2).

(4) The first and last letters are E and E.

There is only one way to arrange those; the remaining 10 letters include 3 R's and 2 S's. The number of arrangements for this case is

10%21%2F%28%283%21%29%282%21%29%29

Add the numbers of arrangements for the four cases to get the final answer.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find the number of ways of arranging all 12 letters of word STRAWBERRIES where
the first and the last letters are vowels.
To make things easier, put the letters of STRAWBERRIES in alphabetical order:

A,B,E,E,I,R,R,R,T,S,S,W

The vowels are A,E,E,I

Case 1. No E's on either end. A and I are on the ends.  
That's 2!=2 ways they can go on the two ends.
Between them are the distinguishable 10-letter arrangements of 
B,E,E,R,R,R,T,S,S,W or 10!/(2!3!2!) 

That's 2![10!/(2!3!2!)] = 302400 ways for Case 1.

Case 2. Exactly 1 E on one end. That puts A or I on the opposite end.
Choose the end (left or right) to put the E on in 2 ways.
That's 2! ways to place the E
Choose letter A or I to put on the opposite end in 2 ways.
That's 2!(2) ways to put the vowels on the ends, exactly one being an E.
Between the vowels on the ends are the distinguishable 10-letter arrangements
of X,B,E,R,R,R,T,S,S,W, or 10!/(3!2!), where the X represents A or I, the one
not chosen for the opposite end.
or 10!/(3!2!)  

That's (2!)(2)[10!/(3!2!)] = 1209600 ways for Case 2.

Case 3. E's on both ends.  
Between them are the distinguishable 10-letter arrangements of
A,B,I,R,R,R,T,S,S,W or 10!/(3!2!) = 302400 for case 3.

For all three cases, that's 302400+1209600+302400 = 1814400 ways.  

Edwin