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stira [4]
3 years ago
6

.!.!.!.!..!.!.!!.!.!..!!..!

Mathematics
1 answer:
frez [133]3 years ago
3 0
J is 126 and k is 27
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How many 3 letter codes are possible using a-j without repeats?
Dennis_Churaev [7]
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
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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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3 years ago
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Nady [450]
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6 0
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