Calculating Permutations of abcdef starting with a,b,c,d and ending with c,d,e,f

[mr_coffee]

Okay so I did a homework problem that was the following:
How many permutations of 5 letters abcde that start with a, b, or c and end with c, d or e.[b/]

To calculate the number of permutations of abcde that start with a, b or c and end with c, d or e, we can use
the Addition Rule to split into disjoint cases before using the Multiplication Rule.
One way to split into disjoint cases is the following.
Case 1: The permutations begin with a or b. Then the first letter can be chosen in 2 ways (a or b), the 5th letter in 3 ways (c, d, or e), and the remaning 3 letters can be placed in the remaining 3 positions in 3!=6 ways. So there are 2 x 3 x 6 = 36 permutations in case 1.

Case 2: The permutation begins with c. Then the first letter can be chosen in 1 way (c), the 5th letter in 2 ways (d or e), and the remaining 3 letters can be placed in the remaining 3 positions in 3! = 6 ways. So there are 1 x 2 x 6 = 12 permutations in case 2. So the answer is:
36 + 12 = 48.

This makes perfect senes to me but when you change the problem alittle:

How many permutations of the six letters abcdef are there in which the first letter is a, b, c or d and the last letter is c, d, e, or f?

Is there a way to do this with only 2 cases or would it have to be 3?

the answer is 288!

 

[quasar987]

To do prob problems, I try to keep things as simple as possible in order to avoid missing things. So I would go at it the robotic way:

answer you're looking for = P({a is the first letter and c is the last} U {a is the first letter and d is the last} U ... U {c is the first letter and e is the last}) = P({a is the first letter and c is the last}) + P({a is the first letter and d is the last}) + ... + P({c is the first letter and e is the last})

Sorry for not answering your specific question.

 

[mr_coffee]

Thats alright thank you, I always forget to try to break it down into a simpler problem such as sets, i'll take another look and apply that!