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aleksandr82 [10.1K]
3 years ago
7

Outside a home, there is a 10-key keypad with the letters A, B, C, D, E, F, G, H, I, and J that can be used to open the garage i

f the correct ten-letter code is entered. Each key may only be used once. How many codes are possible? The number of possible codes is:_____
Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

3,628,800 (aka "10!")

Step-by-step explanation:

The first letter in the code can be 10 possible letters.

Since every letter can only be used once, the second letter can only be 9 possible letters.

E.G.: If the first letter is A, the second letter is B-J.

The third letter would be 8 possible letters, than the fourth would be 7, etc.

In math form, figuring out the possible combinations would be written out as so:

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

An easier way to write this out would be using "10!", which stands for "10 Factorial"

The result of the above equals 3,628,800.

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Well, what's the diffrence from 5/4 to 3/8?

\bf \cfrac{5}{4}-\cfrac{3}{8}\qquad \stackrel{\textit{LCD is 8 clearly}}{\implies }\qquad \cfrac{10-3}{8}\implies \cfrac{7}{8}

so, if we take 5/4 to be the 100%, what is 7/8 off of it in percentage anyway?

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amount&\%\\
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\end{array}\implies \cfrac{\quad \frac{5}{4}\quad }{\frac{7}{8}}=\cfrac{100}{x}\implies \cfrac{5}{4}\cdot \cfrac{8}{7}=\cfrac{100}{x}
\\\\\\
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