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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:

y = - 13, y = 13

Step-by-step explanation:

Given

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Solve the system of equations by substitution: y = x - 2
gizmo_the_mogwai [7]

Step-by-step explanation:

y = x - 2

x + y = 12

Substitute the first expression for y into the second equation.

x + (x - 2) = 12

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2 years ago
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son4ous [18]

Answer:

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Step-by-step explanation:

The fraction is easily converted to one with a denominator of 100 (a power of 10):

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Then the number of interest is ...

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Step-by-step explanation:

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