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Elis [28]
3 years ago
5

How many 3 letter codes are possible using a-j without repeats?

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
5 0
There are 10 letters in the set {a, b, c, d, e, f, g, h, i, j} which is the pool of letters to choose from when making these three letter codes.

We have 10 choices for slot 1
Then 9 choices for slot 2. This is because we can't reuse the choice for slot 1
Then 8 choices for slot 3

Overall, there are 10*9*8 = 90*8 = 720 different permutations

Answer: 720

Note: you can use the nPr permutation formula with n = 10 and r = 3 to get the same answer
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Answer:

  9 ft, 22 ft, 23 ft

Step-by-step explanation:

Let s represent the length of the shortest side. Then the middle length side is (2s+4) and the longest side is (3s-4). The perimeter is the sum of the side lengths:

  54 = s +(2s +4) +(3s -4)

  54 = 6s . . . . . . . . . . . . . . collect terms

  9 = s . . . . . . . . . . divide by 6

  2s+4 = 2·9 +4 = 22

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The lengths of the three sides are 9 feet, 22 feet, and 23 feet.

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(1/2)^2 = 1/4
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OK so:
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Use emilimination to solve 2t 3n=9 and 5t-3n=5
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