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stepan [7]
2 years ago
10

How many 4-digit numbers can be chosen that are even and greater than 3000?

Mathematics
1 answer:
Stella [2.4K]2 years ago
7 0

Answer: 4-digit numbers can be chosen that are even and greater than 3000 is equal to 192.

Step-by-step explanation:

Given data,

4-digit numbers can be chosen that are even and greater than 3000.

For even numbers, each formed 4-digit number must end with 2, 4, or 6. And each such number greater than 3000 must begin with 3, 4, 5, 6, or 7.

Since five digits [1–5] are available, the number may have four or five digits.

All five-digit numbers are greater than 3000, and there are 5! = 120 permutations of the five digits.

Four-digit numbers greater than 3000 must have one of three digits (3 or 4 or 5) as its first digit. The next three digits must be selected from the four remaining digits available, so the number of permutations is: 4 * 3 * 2 = 24.

There are 3 * 24 = 72 numbers of the form 3xxx, 4xxx, 5xxx (all of which are greater than 3000).

The number of possible sequences of [1–5] greater than 3000 include 120 four-digit numbers and 72 five-digit numbers, so the total is 192.

Therefore,

4-digit numbers can be chosen that are even and greater than 3000 is equal to 192.

Learn more about event correlation here: brainly.com/question/24556662

#SPJ9

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<em>Arranging numbers in ascending order: </em>

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<em>Let's count how many times each number occurs in this series of numbers.</em>

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<em>17 </em>+ + +

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<em>21 </em>+ +

<em>22 </em>+

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