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Sagot :
Let's break down the problem into three parts:
### Part 1: Total arrangements of the word "UNIVERSITY"
The word "UNIVERSITY" has 10 distinct letters. To find the total number of different arrangements of these letters, we use the factorial of the total number of letters:
[tex]\[ \text{Total arrangements} = 10! \][/tex]
Given the result, [tex]\( 10! = 3,628,800 \)[/tex].
### Part 2: Arrangements starting with 'U'
If the arrangement must start with 'U', then we only need to arrange the remaining 9 letters: "NIVERSITY". The number of such arrangements is:
[tex]\[ \text{Arrangements starting with 'U'} = 9! \][/tex]
Given the result, [tex]\( 9! = 362,880 \)[/tex].
### Part 3: Arrangements starting with 'U' and not ending with 'Y'
Now, we want the arrangements that start with 'U' but do not end with 'Y'.
1. Fix 'U' at the first position.
2. We have 9 remaining positions and letters.
3. The last position (10th) should not have 'Y'. Thus we have 8 positions where 'Y' can be placed.
4. Arrange the remaining 8 letters in 8 positions.
The number of such arrangements is calculated as the factorial of the remaining 8 letters multiplied by 9 (since 'Y' can take any of the first 9 positions after 'U'):
[tex]\[ \text{Arrangements starting with 'U' and not ending with 'Y'} = 8! \times 9 \][/tex]
Given the result, [tex]\( 8! \times 9 = 362,880 \)[/tex].
### Summary of Results
1. Total arrangements of "UNIVERSITY": 3,628,800
2. Arrangements starting with 'U': 362,880
3. Arrangements starting with 'U' and not ending with 'Y': 362,880
### Part 1: Total arrangements of the word "UNIVERSITY"
The word "UNIVERSITY" has 10 distinct letters. To find the total number of different arrangements of these letters, we use the factorial of the total number of letters:
[tex]\[ \text{Total arrangements} = 10! \][/tex]
Given the result, [tex]\( 10! = 3,628,800 \)[/tex].
### Part 2: Arrangements starting with 'U'
If the arrangement must start with 'U', then we only need to arrange the remaining 9 letters: "NIVERSITY". The number of such arrangements is:
[tex]\[ \text{Arrangements starting with 'U'} = 9! \][/tex]
Given the result, [tex]\( 9! = 362,880 \)[/tex].
### Part 3: Arrangements starting with 'U' and not ending with 'Y'
Now, we want the arrangements that start with 'U' but do not end with 'Y'.
1. Fix 'U' at the first position.
2. We have 9 remaining positions and letters.
3. The last position (10th) should not have 'Y'. Thus we have 8 positions where 'Y' can be placed.
4. Arrange the remaining 8 letters in 8 positions.
The number of such arrangements is calculated as the factorial of the remaining 8 letters multiplied by 9 (since 'Y' can take any of the first 9 positions after 'U'):
[tex]\[ \text{Arrangements starting with 'U' and not ending with 'Y'} = 8! \times 9 \][/tex]
Given the result, [tex]\( 8! \times 9 = 362,880 \)[/tex].
### Summary of Results
1. Total arrangements of "UNIVERSITY": 3,628,800
2. Arrangements starting with 'U': 362,880
3. Arrangements starting with 'U' and not ending with 'Y': 362,880
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