This is a two part question.....
2. How many ways can you arrange the letters A,B,C,D,E without repetition?
3. How many of the arrangements start with a C or end in an A?
To answer these questions, we will use the concept of permutations.
1. How many ways can you arrange the letters A, B, C, D, E without repetition?
For this question, we will calculate the number of permutations without repetition. A permutation is an arrangement of objects in a specific order.
To find the number of ways to arrange the letters without repetition, we need to calculate the factorial of the total number of letters. The factorial of a number is the product of that number and all the positive integers below it.
In this case, we have 5 letters (A, B, C, D, E). Therefore, the number of ways to arrange them without repetition is:
5! = 5 × 4 × 3 × 2 × 1 = 120
So, there are 120 ways to arrange the letters A, B, C, D, E without repetition.
2. How many of the arrangements start with a C or end in an A?
To answer this question, we will consider two separate cases: arrangements that start with C and arrangements that end with A.
Arrangements that start with C:
In this case, we fix C as the first letter, and then we have 4 remaining letters to arrange (A, B, D, E). The number of ways to arrange these 4 letters without repetition is the same as the previous question:
4! = 4 × 3 × 2 × 1 = 24
Arrangements that end with A:
Similarly, we fix A as the last letter, and then we have 4 remaining letters to arrange (B, C, D, E). Again, the number of ways to arrange these 4 letters without repetition is:
4! = 4 × 3 × 2 × 1 = 24
However, it is important to note that some arrangements may satisfy both conditions (start with C and end with A). We need to subtract these arrangements once to avoid double counting.
To find the total number of arrangements that either start with C or end with A, we need to add the arrangements from each case (24 + 24 = 48) and then subtract the duplicate arrangements (which is 1):
Total arrangements = arrangements that start with C + arrangements that end with A - duplicate arrangements
Total arrangements = 24 + 24 - 1 = 47
Therefore, there are 47 arrangements that either start with a C or end with an A.