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.
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
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.
There are no three didgit numbers but there are 63 three digit numbers.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
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
48 of them if digits may not be repeated. 100 if they can.
Any 3 out of 4 = 4. Each set can be arranged 6 ways so 24 different numbers.
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.