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

• mr_coffee
In summary, there are 288 different permutations of six letters abcdef in which the first letter is a, b, c or d and the last letter is c, d, e, or 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?

Last edited:
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})

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!

## 1. How do you calculate the total number of permutations?

To calculate the total number of permutations of abcdef starting with a,b,c,d and ending with c,d,e,f, we use the formula n!/(n-r)! where n is the total number of items and r is the number of items in each permutation. In this case, n=6 and r=4, so the total number of permutations is 6!/(6-4)! = 6!/2! = 6*5*4*3 = 360.

## 2. Can the order of the items in each permutation be changed?

Yes, the order of the items in each permutation can be changed as long as the starting and ending items remain the same. In this case, a,b,c,d and c,d,e,f must remain in the first and last positions, respectively, but the order of the other items can be changed.

## 3. How do you avoid repeating permutations?

To avoid repeating permutations, we use the formula n!/(n-r)! where n is the total number of items and r is the number of items in each permutation. In this case, n=6 and r=4, so we have to divide the total number of permutations by 2! because there are two sets of identical items (c,d). This gives us 6!/2! = 6*5*4*3 = 360, which is the total number of unique permutations.

## 4. Can the starting and ending items be changed?

Yes, the starting and ending items can be changed as long as the other items remain in the same order. For example, if we want to calculate the permutations of abcdef starting with b, c, d, e and ending with b, c, d, e, we would use the same formula as before but with n=5 and r=4. This would give us 5!/1! = 5*4*3*2 = 120 unique permutations.

## 5. Can this formula be applied to other sets of items?

Yes, this formula can be applied to other sets of items as long as the starting and ending items remain the same and there are enough items to create permutations. The formula n!/(n-r)! can be used for any number of items and any number of items in each permutation. It is a fundamental principle in combinatorics and can be applied to a wide range of scenarios.

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