How many positive odd integers less than 500 can be formed from the digits 3, 4, and 5 without repetition?
only the 2 or 3 can be used at the front
without the odd restriction:
number of such number = 2x2x1 = 4
we can actually list them:
345, 354, 435, 453
notice that 1 of them is even
so there are only 3 of them
or
suppose it starts with the 4:
only the two odds are left, so number of such numbers = 1x2x1 = 2
suppose it starts with a 3, so only one odd is left to use in the unit place, leaving the remaining one for the tens place
number of ways = 1x1x1 = 1
total of 3
To find the number of positive odd integers that can be formed from the digits 3, 4, and 5 without repetition, we need to consider a few rules:
1. The last digit must be 3 or 5 since it needs to be odd.
2. The first digit cannot be 0 or 4, as it would result in a number less than 100.
Given these rules, we can proceed step by step to find the answer:
Step 1: Count the possible options for the first digit.
The first digit can be either 3 or 5, so there are 2 options for the first digit.
Step 2: Count the possible options for the second digit.
The second digit can be any of the remaining two digits (either 3 or 5). Since we have used one digit in Step 1, there is only one digit left for the second position. Therefore, there is 1 option for the second digit.
Step 3: Count the possible options for the third digit.
The third digit can be any of the remaining one digit (either 3 or 5). Since we have used two digits in the previous steps, there are no more digits left for the third position. Therefore, there are no options for the third digit.
Step 4: Multiply the options from each step to find the total number of positive odd integers.
To find the total number of positive odd integers, we multiply the number of options from each step:
Total number of positive odd integers = Options for the first digit * Options for the second digit * Options for the third digit
Total number of positive odd integers = 2 * 1 * 0 (since there are no options for the third digit) = 0.
Therefore, there are 0 positive odd integers less than 500 that can be formed from the digits 3, 4, and 5 without repetition.
To find the number of positive odd integers less than 500 that can be formed from the digits 3, 4, and 5 without repetition, we can follow these steps:
Step 1: Determine the possible units place digit.
Since we need the numbers to be odd, the units place digit can only be 3 or 5. The digit 3 appears once, and the digit 5 appears once. Therefore, there are 2 options for the units place digit.
Step 2: Determine the possible tens place digit.
The tens place digit can be any of the remaining two digits (4 and the remaining unused digit from step 1). Therefore, there are 2 options for the tens place digit.
Step 3: Determine the possible hundreds place digit.
The hundreds place digit can be any of the remaining one digit. Therefore, there is 1 option for the hundreds place digit.
Step 4: Multiply the options.
To find the total number of possible numbers, we multiply the number of options for each place value together: 2 x 2 x 1 = 4.
Therefore, there are 4 positive odd integers less than 500 that can be formed from the digits 3, 4, and 5 without repetition.