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To make this simpler, I'll assume that numbers with a leading zero like 0179 are "four digit numbers" for the purposes of this question.

The number of such numbers that can be formed by those numbers without repeating one of the digits is 4x3x2x1 = 24.

We could list them all and add them up, but let's instead just realize that we can choose any digit and put it in any position, and the other digits can then be arranged in 3x2x1 = 6 ways.

So the sum for any position is (0+1+7+9)*6 = 102, and the sum of the set of "four digit" numbers is 102x1000+102x100+102x10+102 = 113322.

You can figure out for yourself what the sum would be if you exclude the numbers that are really three digit numbers (hint: use the technique above to find the sum of the three digit numbers that can be formed using the digits 1, 7, and 9; then subtract that from the other total).

Q: What is the sum of all 4 digit numbers that can be formed using digits 0179no repetition?

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9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.

There are 2000 such numbers.

# 230 # 203 # 320 # 302

With repetition, 94 = 6561. Without repetition, 9*8*7*6 = 3024.

There are 60480 numbers.

Related questions

64 if repetition is allowed.24 if repetition is not allowed.

15 of them.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

125

I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24

Six (6)

There are 2000 such numbers.

9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.

There are 2000 such numbers.

Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77

9*8*7 = 504 of them.

3*2*1 = 6 of them.