3 * 4 * 3 * 2 * 1 = 72;
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.
More than one
18
There are 320 such numbers.
The two-digit numbers with both digits even are formed using the even digits 0, 2, 4, 6, and 8. However, since the first digit (the tens place) cannot be 0, the possible choices for the first digit are 2, 4, 6, and 8 (4 options). The second digit (the units place) can be 0, 2, 4, 6, or 8 (5 options). Therefore, the total number of two-digit numbers with both digits even is (4 \times 5 = 20).
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.
6*6*3 = 108 numbers.
More than one
18
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
Assuming the question refers to the digits 1, 2, ... , 9 the answer is 1344 numbers.
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.
There are 320 such numbers.
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
There are 5760 such numbers.
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
The two-digit numbers with both digits even are formed using the even digits 0, 2, 4, 6, and 8. However, since the first digit (the tens place) cannot be 0, the possible choices for the first digit are 2, 4, 6, and 8 (4 options). The second digit (the units place) can be 0, 2, 4, 6, or 8 (5 options). Therefore, the total number of two-digit numbers with both digits even is (4 \times 5 = 20).