To solve questions related to permutations of the word "SUCCESSFUL," we need to handle the letters and their repetitions carefully. Below are the detailed solutions for each part:
Given:
The word "SUCCESSFUL" has 9 letters with repetitions:
- S: 3 times
- U: 2 times
- C: 2 times
- E: 1 time
- F: 1 time
- L: 1 time
1) There are no restrictions:
To find the total number of different permutations of the word "SUCCESSFUL," we use the formula for permutations of a multiset:
\(\[ \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} \]\)
Here, \(\( n = 9 \)\) (total letters), \(\( n_1 = 3 \)\) (S), \(\( n_2 = 2 \)\) (U), \(\( n_3 = 2 \)\) (C), \(\( n_4 = 1 \)\) (E), \(\( n_5 = 1 \)\) (F), \(\( n_6 = 1 \)\) (L):
\(\[ \frac{9!}{3! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1!} = \frac{362880}{6 \cdot 2 \cdot 2 \cdot 1 \cdot 1 \cdot 1} = \frac{362880}{24} = 15120 \]\)
Answer: 15,120
2) The first letter must be F:
Fixing the first letter as F, we are left with the letters "SUCCESSUL" (8 letters):
- S: 3 times
- U: 2 times
- C: 2 times
- E: 1 time
- L: 1 time
Applying the formula for permutations of this new multiset:
\(\[ \frac{8!}{3! \cdot 2! \cdot 2! \cdot 1!} = \frac{40320}{6 \cdot 2 \cdot 2 \cdot 1} = \frac{40320}{24} = 1680 \]\)
Answer: 1,680
3) The last letter must be U:
Fixing the last letter as U, we are left with the letters "SUCCESSFL" (8 letters):
- S: 3 times
- U: 1 time
- C: 2 times
- E: 1 time
- F: 1 time
- L: 1 time
Applying the formula for permutations of this new multiset:
\(\[ \frac{8!}{3! \cdot 2! \cdot 1! \cdot 1! \cdot 1!} = \frac{40320}{6 \cdot 2 \cdot 1 \cdot 1 \cdot 1} = \frac{40320}{12} = 3360 \]\)
Answer: 3,360
4) The first letter must be S AND the last letter must be S:
Fixing both the first and last letters as S, we are left with the letters "UCCESSFU" (7 letters):
- S: 1 time
- U: 2 times
- C: 2 times
- E: 1 time
- F: 1 time
- L: 1 time
Applying the formula for permutations of this new multiset:
\(\[ \frac{7!}{1! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1!} = \frac{5040}{1 \cdot 2 \cdot 2 \cdot 1 \cdot 1 \cdot 1} = \frac{5040}{4} = 1260 \]\)
Answer: 1,260
5) The first letter must be E OR the last letter must be L:
We'll find each case separately and subtract the intersection (both conditions simultaneously), to avoid double-counting.
Case 1: First letter is E:
Fixing the first letter as E, we are left with the letters "SUCCESSFUL" (8 letters):
- S: 3 times
- U: 2 times
- C: 2 times
- F: 1 time
- L: 1 time
\(\[ \frac{8!}{3! \cdot 2! \cdot 2! \cdot 1!} = 1680 \]\)
Case 2: Last letter is L:
Fixing the last letter as L, we are left with the letters "SUCCESSFU" (8 letters):
- S: 3 times
- U: 2 times
- C: 2 times
- E: 1 time
- F: 1 time
\(\[ \frac{8!}{3! \cdot 2! \cdot 2! \cdot 1!} = 1680 \]\)
Intersection: First letter is E AND last letter is L:
Fixing both as E and L, we are left with "SUCCESSF" (7 letters):
- S: 3 times
- U: 1 time
- C: 2 times
- F: 1 time
\(\[ \frac{7!}{3! \cdot 2! \cdot 1! \cdot 1!} = \frac{5040}{6 \cdot 2 \cdot 1 \cdot 1} = \frac{5040}{12} = 420 \]\)
Combining these results:
\(\[ \text{Total} = \text{Case 1} + \text{Case 2} - \text{Intersection} = 1680 + 1680 - 420 = 2940 \]\)
Answer: 2,940
