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How many 2 digit numbers can be formed from 0 to 9?
Oh, dude, there are 90 two-digit numbers you can make from 0 to 9. You've got 10 options for the first digit (0-9) and 9 options for the second digit (0-9 excluding the one you already picked for the first digit). So, like, 10 times 9 equals 90. Math can be fun when you're making numbers do the cha-cha.
How do you multiply by 2 or 3 digit numbers?
first digit time second digit and second digit times first digit then repeat
How many possible 5 digit combinations are there using 5 digits?
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
How many odd numbers with 4 distinct digits can be formed using the digits in 234567?
The last digit must be odd: three possibilities. The other digits can be anything else, as long as they're different: five times four times three possibilities. Thus, the answer is 3 x 5 x 4 x 3 = 180.
How many times does the digit 6 appear in the list of all integers from 1 to 100?
10 times for the one's digit, 1-100 10 times for the ten's digit, 60-70 = 20 times
How many 6 digit numbers can be formed that contains no zeros two times the digit 1 two times the digit 2?
16000
How many 4 digit numbers can be formed using 1 2 3 4 5 6 7 8 and 9 as many times as possible?
6,561
How many 5 digit numbers containing two times the digit 6 can be formed from the digits 123456?
No digit that is 5 places long is equal to 6 times 2 other than 12.000
How many 2-digits numbers can be formed from the digits 563 and 2?
To form 2-digit numbers from the digits 5, 6, 3, and 2, we can use each digit as both the tens and units place. The possible combinations for the tens place are 5, 6, 3, and 2, giving us 4 options. For each choice of the tens digit, we have 4 choices for the units digit as well. Therefore, the total number of 2-digit numbers that can be formed is (4 \times 4 = 16).
How many three digit or four digit even numbers can be formed from the set (23567)?
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.
How many 3 digits even numbers can be formed if repetition of digits is possible?
To form a 3-digit even number, the last digit must be an even digit (0, 2, 4, 6, or 8), giving us 5 options for the last digit. The first digit can be any digit from 1 to 9 (9 options), and the second digit can be any digit from 0 to 9 (10 options). Therefore, the total number of 3-digit even numbers is calculated as (9 \times 10 \times 5 = 450).
How many digit numbers can be formed from the digit 2 3 5 7 8?
To determine how many digit numbers can be formed using the digits 2, 3, 5, 7, and 8, we need to consider the number of digits in the numbers we are forming. For a 1-digit number, we can use any of the 5 digits. For a 2-digit number, we can choose 2 out of the 5 digits and arrange them, giving us (5 \times 4) combinations. We can continue this for 3-digit, 4-digit, and 5-digit numbers, which will yield (5), (20), (60), and (120) respectively. Therefore, the total number of digit numbers is (5 + 20 + 60 + 120 = 205).
How many 2 digit numbers can be formed from 0 to 9?
Oh, dude, there are 90 two-digit numbers you can make from 0 to 9. You've got 10 options for the first digit (0-9) and 9 options for the second digit (0-9 excluding the one you already picked for the first digit). So, like, 10 times 9 equals 90. Math can be fun when you're making numbers do the cha-cha.
How many two digit numbers have both of their digit even?
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).
How many 3 digit numbers can be formed from the digits 1 2 3 4 5 6 7 8 9 if each digit is used only once?
Multiply the number of possible starting numbers by the number of possible middle numbers by the number of possible end numbers to get your result.In example, The possible starting numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let's say I picked nine as the starting number. Since your question states that each number can only be used once, we eliminate nine from the selection of middle and end numbers. Now, the choices for the possible middle and end numbers are 1, 2, 3, 4, 5, 6, 7, and 8.Possible starting numbers= 9Possible middle numbers= 8Multiply 9 by 8. You get 72 different choices for a two digit number.Let's say that the middle number I picked was two. We then remove the number two from the possible choices for the final numbeselections: 1, 3, 4, 5, 6, 7, and 8.Possible outcomes for two digit number= 72 (which is 9 times 8)Possible end numbers = 7Multiply 72 by 7 to get the possible outcomes for a three digit number with each digit used only once.72 times 7 = 504. You have 504 possible outcomes.
How many four digits can be formed in 1 2 3 4 5 and 6 without repeating the numbers?
To form a four-digit number using the digits 1, 2, 3, 4, 5, and 6 without repeating any digits, you can choose the first digit in 6 ways, the second digit in 5 ways, the third digit in 4 ways, and the fourth digit in 3 ways. Therefore, the total number of four-digit combinations is calculated as (6 \times 5 \times 4 \times 3 = 360). Thus, 360 different four-digit numbers can be formed.
How many numbers can be formed using 2 3 4 and 5 as many times as possible?
there are 22
