Respuesta :

konrad509

Let's calculate first, how many such numbers there are, where the same three digits are different form 0.


[tex] \underbrace{1\cdot1\cdot1\cdot1}_{\text{XXX0}}\cdot\underbrace{9}_{\text{X=\{1,2,\ldots,9\}}}\cdot\underbrace{3}_{\text{XXX0,XX0X,X0XX}}=27[/tex]


Now, the numbers where there are three 0's. There are only 9 of those

[tex] 1000,2000,\ldots,9000 [/tex]


So, there are [tex] 27+9=36 [/tex] such numbers.




facundo3141592

We must count how many four-digit positive integers have exactly three same digits and at least one zero.

We will see that are 36 numbers that meet these conditions.

The easier way to find this is if we just count.

For a four-digit number we can start with:

1000

We have a four-digit number, the zero repeats 3 times, and we have at least one zero, so this meets the conditions.

Similarly, we can construct:

1110

1101

1011

In the 3 cases, we have 3 same digits and at least one zero. At the moment we have 4 numbers that meet the conditions.

With  a similar approach we can see that:

2000

2220

2202

2022

Also meet the conditions, so we will see that for each digit {1, 2, ..., 9} we can construct 4 four-digit numbers that meet the criterion.

Then the total number of four-digit positive integers that have exactly three same digits and at least one zero is:

9*4 = 36

There are 36 of these numbers.

If you want to learn more, you can read:

https://brainly.com/question/17015120