Respuesta :
In probability, problems involving arrangements are called combinations or permutations. The difference between both is the order or repetition. If you want to arrange the letters regardless of the order and that there must be no repetition, that is combination. Otherwise, it is permutation. Therefore, the problem of arrange A, B, C, D, and E is a combination problem.
In combination, the number of ways of arranging 'r' items out of 'n' items is determined using n!/r!(n-r)!. In this case, you want to arrange all 5 letters. So, r=n=5. Therefore, 5!/5!(505)! = 5!/0!=5!/1. It is simply equal to 5! or 120 ways.
In combination, the number of ways of arranging 'r' items out of 'n' items is determined using n!/r!(n-r)!. In this case, you want to arrange all 5 letters. So, r=n=5. Therefore, 5!/5!(505)! = 5!/0!=5!/1. It is simply equal to 5! or 120 ways.
Answer:
120 ways
Explanation:
Since there are no repeating letters, and there are 5 total letters, there are 5!= 120 ways to arrange them. In other words, there are 5 slots to place the first letter in, then 4 slots for the second letter, 3 for the third, 2 for the fourth, and then the last letter goes in the 1 slot that is left.
I hope this is correct and if it is I hope it helps. I hope you have a great day and remember it doesn't matter what other people say or think about you, you are still and always will be strong, smart, and an amazing person. Have a wonderful day.
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