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madam [21]
2 years ago
7

If a pin consists of 5 numbers , how many different PIN number codes are possible ?

Mathematics
1 answer:
denis-greek [22]2 years ago
6 0

Answer:

The logical way is very simple:

Since you know that there are 5 digits and each must be different, start with the first digit and work your way up.

The first digit can be any of 10 digits (0–9).

However, the second digit can be any of only 9 digits as 1 digit has already been used and cannot be repeated.

Similarly, applying the same logic, there are only 8 options for the third digit, 7 for the fourth digit, and 6 for the fifth and final digit.

Now, multiply all these numbers together to get the total number of configurations like so:

10⋅9⋅8⋅7⋅6=30240

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COME ON ITS 1 QUESTION FOR 50 POINTS!
ale4655 [162]

Answer:

a = 3

b = 2

c = 0

d = -4

Step-by-step explanation:

Form 4 equations and solve simultaneously

28 = a(2)³ + b(2)² + c(2) + d

28 = 8a + 4b + 2c + d (1)

-5 = -a + b - c + d (2)

220 = 64a + 16b + 4c + d (3)

-20 = -8a + 4b - 2c + d (4)

(1) + (4)

28 = 8a + 4b + 2c + d

-20 = -8a + 4b - 2c + d

8 = 8b + 2d

d = 4 - 4b

Equation (2)

c = -a + b + d + 5

c = -a + b + 4 - 4b+ 5

c = -a - 3b + 9

28 = 8a + 4b + 2c + d (1)

28 = 8a + 4b + 2(-a - 3b + 9) + 4 - 4b

28 = 6a - 6b + 22

6a - 6b = 6

a - b = 1

a = b + 1

220 = 64a + 16b + 4c + d (3)

220 = 64(b + 1) + 16b + 4(-b - 1 - 3b + 9) + 4 - 4b

220 = 60b + 100

60b = 120

b = 2

a = 2 + 1

a = 3

c = -3 - 3(2) + 9

c = 0

d = 4 - 4(2)

d = -4

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