# Pls asap for 20 points i will mark correct brainle est might be easy. how many 4-digit

Pls asap for 20 points i will mark correct brainle est might be easy. how many 4-digit numbers can you write if you cannot use any other digits except 0 and 2?

## This Post Has 8 Comments

1. que81hhh says:

Think of the four digit number like a slot. Each slot has 3 different possibilities so that makes the total slot have 3*3*3*3 probabilities, so 81

2. kaylaelaine18 says:

Hope this helped!!!XD

3. Expert says:

i'm pretty sure the answer is a. decreases with increase in

4. emaliemcfalls1234 says:

You can write the combination formula which is in the picture. Now if you can't guess what the varibles mean I'll tell you. N is the amount of digits you're allowed to use, and k is the amount of spots allowed in a number. Plus if the factorial sign or ! isn't part of your common knowledge, it means to multiply all  below it stopping at 1. Finally if you comput your information into the formula you find out that there's 24 combinations. I hope this answers your question.

$Pls asap for 20 points i will mark correct brainle est might be easy. how many 4-digit numbers can$

5. Expert says:

table 3 represents a direct variation function. : )

$Which table represents a direct variation function? ​$

6. xxcecccc says:

1111      2222

1122    2211

1222   2111

2221   1112

1221    2112

1212    2121

I believe that the answer is 12 possible 4-digit numbers.

You have a choice of only 3 digits to fill in 4 places.
You can place any of 3 digits in each of the 4 places, so you get
3 * 3 * 3 * 3 = 81

8. jackkie10 says:

8

Steps:

Since the four digit number cannot start with a 0, it has to start with a 2. So we are left with determining the number of ways to fill the rest 3 digits using only 0's or 2's. 2xxx.

For the third digit position, there are 2 choices (0 or 2) , for each of these, there are 2 choices at the second digit, and for each such choice there are 2 possible choices for the first digit. The total of possibilities is 2*2*2=2^3=8.

Interestingly, this is the same number of possibilities encoded in 3 bits in the binary system.