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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= 9
Possible middle numbers= 8
Multiply 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 = 7
Multiply 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.
What else can I help you with?
How many 4 digit numbers can be formed from digits 123456789?
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
How many 4 digit numbers can be formed by the digits 1 2 3 4 using each digit only once?
24 different numbers.
How many digits numbers each less than 500 can be formed from the digits 13467 if repetitions is allowed?
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
How many four digit numbers can be formed out of 123 if a digit can be repeated?
There are 3 values (1, 2, 3) for each of the 4 digits. Therefore, there are 3*3*3*3 or 81 four digit numbers that can be formed.
How many even 4 digit numbers can be formed from digits 2 6 8 9 if each digit can only be used once?
18
How many 4 digit numbers have 6 as their second digit?
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
How many 3 digit numbers can be formed from the digits 1 2 3 4 5 6 if each digit is used only once?
20
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 6 digit numbers can be formed with 6 digits?
If the digits may not be repeated, there are 6x5x4x3x2x1 = 720 numbers, If digits can be repeated, the answer is 6x6x6x6x6x6 = 46,656 numbers If zero is ong the available digits but is excluded from the first place. 5x6x6x6x6x6=38,880 numbers
How many different three digit numbers can be formed from the digits 0 through 9 if the first digit must be odd and the last digit must be even?
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
What is the sum of the greatest and smallest four digit numbers formed by using each of d digits 6039 atleast once?
9630
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.
