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Daniel [21]
2 years ago
13

How many different four-digit numbers can be formed with the numbers 7; 4; 5; 1; 2; 9; and 8?

Mathematics
2 answers:
ExtremeBDS [4]2 years ago
6 0

Answer:

I believe it's 2401.

Step-by-step explanation:

7 x 7 x 7 x 7 = 2401

salantis [7]2 years ago
4 0

Answer:

2 answers.

Step-by-step explanation:

ANSWER 1: With Repetition

Numbers: 1,2,4,5,7,8

There are 4 digits __ __ __ __

1st digit can be filled by any of the 6 numbers.

Similarly 2nd, 3rd and 4th digits.

Hence each digit will have 6 possibilities.

Therefore no.of 4 digit numbers that can be formed are 6 x 6 x 6 x 6 = 1296

ANSWER 2: Without Repetition

4 digits __ __ __ __

Now the first digit can be filled by any of the six numbers. Therefore there are 6 possibilities

The second digit will have only 5 possibilities as one of the numbers gets used up by the 1st digit.

Similarly the 3rd and 4th digits will have 4 and 3 possibilities respectively.

Therefore no.of 4 digit numbers that can be formed are 6 x 5 x 4 x 3 = 360

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Answer:

<u>The sum of the first 5 terms of the series is 77.</u>

Step-by-step explanation:

1. Let's review the information provided to us for solving the sum of the first 5 terms of the series:

1st term = 7

2nd term = - 14

3rd term = 28

4th term = - 56

2. Let's solve the sum of the first 5 terms of the series:

Aₙ = Aₙ₋₁ * -2 ; A₁ = 7

A₁ = 7

A₂ = A₁ * -2 = 7 * - 2 = - 14

A₃ = A₂ * - 2 = -14 * - 2 = 28

A₄ = A₃ * - 2 = 28 * - 2 = - 56

A₅ = A₄ * - 2 = -56 * - 2 = 112

Sum of the first 5 terms of the series = 7 - 14 + 28 - 56 + 112

<u>Sum of the first 5 terms of the series = 77</u>

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Step-by-step explanation:

I've done this before, very easy.

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