What would you like to do?

# Can you use a digital copy more than once?

# How many 4-digit numbers can be formed by the digits 1 2 3 4 using the digits more than once in each number example 1112 1122?

4 4 = 256.
But, if you want to actually exclude those numbers which happen not to include repeated digits (e.g. 4321), then.
4 4 - 4! = 232

# Greatest six digit even number no digit used more than once?

Use the largest available digit possible starting with the hundredthousands digit, making the ones digit the largest even digit {0,2, 4, 6, 8} that has not been used. The req…uired number is: 987654

# How many 4-digit numbers can be formed by the digits 01 2 3 456789 using the digits more than once in each number example 1112 1122?

Four digit numbers span from 1000 to 9999..
This totals (9999-1000) +1 = 9000 different four digit numbers

# Explanation for the answer to how many three digit prime numbers can be formed using only odd prime digits and digits may be used more than once?

The odd prime digits are 3, 5, and 7. but the final digit of your three-digit prime number can't be 5. So there are only 18 combinations to check. You can list these numbers a…nd start going through them to see which are prime. The easiest test is divisibility by 3 (sum the digits) and that quickly eliminates 7 of them. You are left with: 337 353 373 533 553x 557 577 733 737x 757 773 Of these, only two are not prime. I marked them with an "x". I used a table of primes to find these and don't know an easier way.

# How many passwords can be made from 3 letters followed by 3 digits if no letter or digit can be used more than once?

3!*3!=36 For example, if you want to use the alphabets a,b,c and three digits 1,2,3 you can make the following 36 different passwords ( 3 letters followed by 3 digits ) abc…123 abc132 abc231 abc213 abc312 abc321 acb123 acb132 acb231 acb213 acb312 acb321 bca123 bca132 bca231 bca213 bca312 bca321 bac123 bac132 bac231 bac213 bac312 bac321 cab123 cab132 cab231 cab213 cab312 cab321 cba123 cba132 cab231 cab213 cab312 cab321

# How many numbers greater than 4000 can be formed with the digits 2 3 4 5 and 6 if no digit is used more than once in a number?

The numbers between 4,000 and 10,000 _ _ _ _ 1st place only can filled by 4, 5, and 6 2nd place only can filled by the other 4 digit 3rd place only can filled by the… other 3 digit 4th place only can filled by the other 2 digit So: 3 x 4 x 3 x 2 = 72 numbers The numbers greater than 10,000 5 x 4 x 3 x 2 x 1 = 120 numbers The total is 72 + 120 = 192 numbers can formed.

# How many 5 digit numbers that are divisible by 4 can be formed using the digits from 0 to 6 if no digit is to occur more than once in each number?

There are 7 digits from 0-6 To determine if a number is divisible by 4, we only need look at the last 2 digits. Possibilities are: [135][26] => 12, 16, 32, 36, 52, 56 a…nd [0246][04] => 04, 20, 24, 40, 60, 64 (ignoring 00 and 44) There are 8 combinations without a 0, and 4 with a 0. Of the 8 combinations without a 0, We can only pick 4 of the remaining digits for the ten thousand digit. This leaves us with 4 digits for the thousands digit, 3 for the hundreds digit Of the 4 combinations with a 0, similarly we have 5 X 4 X 3 possibilities for the rest of the digits. 8 * 4 * 4* 3 + 4 * 5 * 4 * 3 = 32 * 12 + 20 * 12 = 52 * 12 = 624 There are 624 possible numbers.

# How many ways are there to write a 3-digit positive integer using digits 1 3 5 7 and 9 if no digit is used more than once?

There are 5 digits, and if none can be repeated then the first digit can be chosen in 5 ways, the second can be chosen in 4 ways and the third in 3. 5 x 4 x 3 = 60.

# How many 2-digit numbers can be formed from the digits 4 6 and 8 Assume no number can be used more than once?

Six. 46 48 64 68 84 86

# How many 3 digit numbers can you make using the digits 1 3 5 and 7 by using the digits more than once?

Using the combination C(4,3) I got the answer 24. Its either a combination or a permutation.

Answered

# How many even 3-digit numbers can be formed from digits 1 3 5 and 7 if each digit can be used more than once?

If your only option is to assemble the digits into a single number, then the answer is 0, as you need at least one even digit to make an even number, and the digits provided a…re all odd ones. If on the other hand, you can use operators on them (e.g. 1 + 3 + 5 + 7), then you actually have quite a large number of possibilities. For example: 1 + 3 + 5 + 7 - 1 + 3 + 5 + 7 1 - 3 + 5 + 7 (1 + 3) / ( 7 - 5) 1 + â(7 + 5 - 3) 3 1 + 5 7 etc. etc. Alternatively, if you're not worried about them being even numbers, then the answer is 24.

Answered

# How many 2 digits numbers can you have using 0 through 9 if you can use the numbers more than once?

Assuming that 00, 01, 02, etc. count as two digit numbers, then there are one hundred (1 through 99 and including 0). If you are not allowed to count 00, 01, 02, 03, 04, 0…5, 06, 07, 08, or 09 because they simplify to one digit numbers, then you have ten less--so a total of ninety two digit numbers.

Answered

# How many different 3 digit identification cards can be made is the digits can be used more than once. if the first digit must be a 5 and repetitions are not permitted?

All the numbers from 500 to 599 EXCEPT 500, 511, 522, 533, 544, 550 - 559, 566, 577, 588 & 599. That is 100 less 19 ie 81. If you regard 505 as being a repetition then there a…re 9 less overall.

Answered

# How many ways are there to write a 3 digit positive integer using digits 1 3 5 7 and 9 if no digit is used more than once?

This is a question of permutations; the answer is equal to the factorial of 5 (number of digits) divided by the factorial of 3 (number used in each selection), written 5! / 3!…. This equals 120 / 6, or 20 ways.