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professor190 [17]
1 year ago
14

a password system uses four digits from 0 to 9. how many different four-digit passwords with no digit repeated are possible?

Mathematics
1 answer:
JulijaS [17]1 year ago
3 0

Total 5040 passwords can be generated when no digits are repeated

• Permutation in mathematics is an arrangement of objects in an definite order. The elements or sets or things are arranged in sequential order or linear order.

• Permutation is classified into four types :

1) where repetition is not allowed

2) where repetition is allowed

3) objects that are non distinct  

4) circular permutation

• The formula for calculating permutation is n!/(n-r)!

As we are given that the password uses four digits from 0-9 and repetition is not allowed.

So, n = 10 and r = 4

Using the formula for permutation we get

= 10! / (10-4)!

= 10!/6!

= 10 * 9 * 8 * 7 * 6! / 6!

= 10 * 9 * 8 * 7

= 5040

Therefore, 5040 four digit passwords are possible when no digit is repeated.

Learn more about permutaion here

brainly.com/question/12468032

#SPJ4

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Ahh yes, negative exponents always give us a scare once and a while. All the negative means is to flip the position of its base. For instance, if x has a negative exponent and x in the denominator, all you would have to do is move x to the numerator with the same power (except it's no longer negative). Before we substitute x and all the other variables which the values given, let's eliminate the negatives first.

After flipping positions/eliminating the negative exponents it should look like this:

\frac{ 2^{2} x^{3} }{ y^{5} }
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\frac{ 4 x^{3} }{ y^{5} }
now that everything is simplified, and all negative exponents are eliminated we can substitute x with 2, and y with (-4).
\frac{4 (2)^{3} }{ (-4)^{5} }
which simplifies to 
\frac{4(8)}{(-1024)} = \frac{32}{-1024} = - \frac{1}{32}

Final Answer: - \frac{1}{32} [/tex]
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