To solve this problem, we need to use the concept of permutations in combinatorics. Here are the steps:

1. The word "PERMUTATIONS" has 12 letters in total. However, since we want the words to start with 'P' and end with 'S', we are left with 10 letters to arrange.

2. Among these 10 letters, we have some repetitions. The letter 'T' appears twice, and the letter 'A' appears twice.

3. The number of ways to arrange 'n' distinct items is given by 'n!'. However, if there are repetitions, we need to divide by the factorial of the number of repetitions.

4. So, the number of ways to arrange the 10 letters 'E', 'R', 'M', 'U', 'T', 'T', 'I', 'O', 'N', 'A', 'A' is 10! / (2! * 2!).

5. Therefore, the letters of the word "PERMUTATIONS" can be arranged in 10! / (2! * 2!) ways if the words start with 'P' and end with 'S'.

Question

To solve this problem, we need to use the concept of permutations in combinatorics. Here are the steps:

1. The word "PERMUTATIONS" has 12 letters in total. However, since we want the words to start with 'P' and end with 'S', we are left with 10 letters to arrange.

2. Among these 10 letters, we have some repetitions. The letter 'T' appears twice, and the letter 'A' appears twice.

3. The number of ways to arrange 'n' distinct items is given by 'n!'. However, if there are repetitions, we need to divide by the factorial of the number of repetitions.

4. So, the number of ways to arrange the 10 letters 'E', 'R', 'M', 'U', 'T', 'T', 'I', 'O', 'N', 'A', 'A' is 10! / (2! * 2!).

5. Therefore, the letters of the word "PERMUTATIONS" can be arranged in 10! / (2! * 2!) ways if the words start with 'P' and end with 'S'.

Knowee AI · Accepted Answer

To solve this problem, we need to use the concept of permutations in combinatorics. Here are the steps:

1. The word "PERMUTATIONS" has 12 letters in total. However, since we want the words to start with 'P' and end with 'S', we are left with 10 letters to arrange.

2. Among these 10 letters, we have some repetitions. The letter 'T' appears twice, and the letter 'A' appears twice.

3. The number of ways to arrange 'n' distinct items is given by 'n!'. However, if there are repetitions, we need to divide by the factorial of the number of repetitions.

4. So, the number of ways to arrange the 10 letters 'E', 'R', 'M', 'U', 'T', 'T', 'I', 'O', 'N', 'A', 'A' is 10! / (2! * 2!).

5. Therefore, the letters of the word "PERMUTATIONS" can be arranged in 10! / (2! * 2!) ways if the words start with 'P' and end with 'S'.

In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S?

Question

In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S?

Solution

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