Write the number of permutations in factorial form. Then simplify. J K L M N O P A.7!; 720 B.7!; 49 C.7!; 5,040 D.8!: 40,320

Accepted Solution

Answer:The number of permutations of the group of letters given (J, K, L, M, N, O and P) is 7! = 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5,040. Step-by-step explanation:The number of permutations without repetition is the number of possible sequences that can occur if we take the letters one by one and is calculated with the factorial of the total number of letters in the group. Before starting we have 7 possibilities, one for each letter. Once the first letter is chosen randomly, we will have 6 possibilities, after taking the third letter we will have 5 possibilities, and so on. In total the total number of possibilities to obtain the 7 letters in a certain order are 7 by 6 by 5 by 4 by 3 by 2 by 1, sayed in other words factorial of 7 (7!). The result of the multiplication gives 5,040. In short we can get the seven letters in 5,040 different sequences.