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?

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  1. 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. 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.

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

  3. table 3 represents a direct variation function. : )

    [tex]Which table represents a direct variation function? ​[/tex]

  4. 1111      2222

    1122    2211

    1222   2111

    2221   1112

    1221    2112

    1212    2121

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

  5. 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

  6. 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.

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