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.
The answer is e=289. You just minus 1201 by 912 to get 289.
7 tenths is 0.7. The fraction form is 7/10, since the digit of 7 is in the tenths place.
I hope this helps, please Brainliest me, and have a great day! :D
1.)Substitute using given information: 8x-2(3)-3
2.)Simplify: 8x-6-3=8x-9
8x-9
