# Thread: How many 4 digit numbers can you make

1. ## How many 4 digit numbers can you make

Hi, I really need help with this
How many odd, 4 digit numbers can you make out of 0, 3, 7 and 8? I am so bad at math...

2. This would be a permutation question.
We have 4 choices, 0, 3, 7, 8
As we wish to make a 4-digit number abcd.

For d, the units, we have 2 choices, 3 or 7.
And for each of the other digits, a,b,c we have 4 choices, 0,3,7,8.

Hence the our total number of choices are 4x4x4x2

3. Thank you!

4. If any of the digits 0, 3, 7, and 8 can be repeated in the 4-digit number, then the answer is 196.

If none of the digits 0, 3, 7, and 8 can be repeated in the 4-digit number, then the answer is 8.

5. I'm pretty sure it would just be 4! (that's factorial not an exclamation).

Oh and 4! = 24.

6. Based on the wording of the question, there is more than one possible answer.

A list of all 4-digit numbers using only the digits 0, 3, 7, and 8 with repeating any digit more than once is:

3003 3007 3033 3037 3073 3077 3083 3087 3303 3307 3333 3337 3373 3377 3383 3387 3703 3707 3733 3737 3773 3777 3783 3787 3803 3807 3833 3837 3873 3877 3883 3887 7003 7007 7033 7037 7073 7077 7083 7087 7303 7307 7333 7337 7373 7377 7383 7387 7703 7707 7733 7737 7773 7777 7783 7787 7803 7807 7833 7837 7873 7877 7883 7887 8003 8007 8033 8037 8073 8077 8083 8087 8303 8307 8333 8337 8373 8377 8383 8387 8703 8707 8733 8737 8773 8777 8783 8787 8803 8807 8833 8837 8873 8877 8883 8887

There are 96 numbers in that list.

A list of all 4-digit numbers using only the digits 0, 3, 7, and 8 without repeating any digit more than once is:

3087 3807 7083 7803 8037 8073 8307 8703

There are 8 numbers in that list.

7. Originally Posted by I-Think
This would be a permutation question.
We have 4 choices, 0, 3, 7, 8
As we wish to make a 4-digit number abcd.

For d, the units, we have 2 choices, 3 or 7.
And for each of the other digits, a,b,c we have 4 choices, 0,3,7,8.

Hence the our total number of choices are 4x4x4x2
You were on the right track, however a 4 digit number cant start with a 0.
That is:
Say, we wish to make a 4-digit number abcd.
We want a 4-digit number, so a cannot be 0. So we have only 3 choices for a.

Hence the our total number of choices are 3x4x4x2

Originally Posted by JSB1917
I'm pretty sure it would just be 4! (that's factorial not an exclamation).
Oh and 4! = 24.
Hey JSB1917,
Why do you think it is 4!? that would be counting all permutations of the letters of the string 0378. But we need the total number of "odd" "numbers" in such strings. Note that 0378 itself is even

8. Thank you for all of your posts and help.
Today in math class we got to this exercise again, tho a bit different example. By the way, turns out that I forgot to mention a detail about that exercise I gave in my initial post - the numbers can't be repeated.
So anyways here's what exercise we solved today:
How many even, 4 digit numbers, without repeating the digits can you make out of 0, 2, 7 and 9?
Here's what the teacher wrote on the board for this one:

So then she wrote all the possible combinations, which there were 18, and then we chose all numbers that were even, and we got only 10 numbers. So the answer is that you can make 10 even, 4 digit numbers, without repeating the digits out of 0, 2, 7 and 9 .

### how many 4 digit combinations with 10 numbers

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