Hi Corban,

I am not going to use the term combinations since that is a technical term in mathematics, instead I am going to use to the more generic term arrangements.

In your list you include 137, 731 and 317 as different arrangement so I assume 173, 713 and 371 are different arrangements also. You don't see to allow the same digit to appear in the same arrangement, so for example 773, 757 and 333 are not allowed. If I am correct in my assumptions then mathematicians would say you are looking for the number of permutations of 7 items taken 3 at a time.

If you are going to list all the permutations then you have 7 choices for the leftmost digit. Since you can't repeat digits you now have 6 choices for the middle digit. Thus you have $7 \times 6 $ choice for the first two digits. At this point you have only 5 choices for the rightmost digit. Thus you have $7\times 6 \times 5$ ways to construct a 3 digit permutation from the 7 digits from 1 to 7.

I hope this helps,
Penny

Corban replied

In your answer you assumed that numbers could not be repeated in sequence. I need to know how many combinations there are using 3 digits and the numbers 1-7 Every possible combination. 777 127 177 656 etc there are no exclusions apart from using the numbers 0 9 8 I was told the equation is 7x7x7 but its driving me insane because I don't think that's right..

Please help
Thanks

Hi again Corban,

Yes, the answer is $7 \times 7 \times 7.$ As I did above, imagine that the 3 digit arrangements are 3 digit numbers. To list the all start with the leftmost digit. You have 7 choices so your possible one digit numbers are

• 1
• 2
• 3
• 4
• 5
• 6
• 7

Now the second digit. Each of the above one digit numbers can be extended to a two digit number by appending any of the digits 1 to 7 in the second place. Thus all possible two digit numbers are

• 11
• 12
• 13
• 14
• 15
• 16
• 17
• 21
• 22
• 23
• .
• .
• .
• 71
• 72
• 73
• .
• .
• 77

I didn't write them all but you can see that since each one digit number can be extended to a two digit number in 7 ways there are $7 \times 7$ two digit numbers.

Finally the third digit. Each of the two digit numbers can be extended to a three digit number by appending and of the digits 1 to 7 in the third place. Hence all possible three digit numbers are

• 111
• 112
• 113
• 114
• .
• .
• .
• 117
• 121
• 122
• 123
• .
• .
• .
• 771
• 772
• 773
• .
• .
• .
• 777

Again I didn't list them all but you can see that since each two digit number can be extended to a three digit number in 7 ways there are $7 \times 7 \times 7 $ three digit numbers.

Penny