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- https://stats.libretexts.org/Sandboxes/JolieGreen/Finite_Mathematics_-_Spring_2023_-_OER/10%3A_Sets_and_Counting/10.03%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. \[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
- https://stats.libretexts.org/Under_Construction/Purgatory/FCC_-_Finite_Mathematics_-_Spring_2023/10%3A_Sets_and_Counting/10.03%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. \[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
- https://stats.libretexts.org/Courses/Fresno_City_College/New_FCC_DS_21_Finite_Mathematics_-_Spring_2023/10%3A_Sets_and_Counting/10.03%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. \[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
