48 of them if digits may not be repeated. 100 if they can.
Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
The smallest three-digit number divisible by the first three prime numbers (2, 3, and 5) and the first three composite numbers (4, 6, and 8) is 120.
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).
512
from 100 to 999 it is 900
27 three digit numbers from the digits 3, 5, 7 including repetitions.
Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
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
Any 3 out of 4 = 4. Each set can be arranged 6 ways so 24 different numbers.
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
Choose from 4 for the first number, 3 for the second number, and 2 for the third number. Therefore there are 4*3*2 = 24 three digit numbers can be formed from 3769 if each digit is used only once.
24 of them. So we have a 2 or 6 in the unit position, therefore (2)(4)(3) = 24 even three-digit numbers, can be formed.
For the first digit there would be 4 options For the second digit there would be 3 options for the third digit there would be 2 options Thus 4*3*2 = 24 three digit numbers can be formed if each digit can only be used once in a 3 digit number made.
To form a three-digit number using the digits 1-7, we can choose any of the 7 digits for each of the three places (hundreds, tens, and units). Therefore, the total number of 3-digit combinations can be calculated as (7 \times 7 \times 7), which equals 343. Thus, there are 343 different three-digit numbers that can be formed using the digits 1-7.