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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.
How many 4 digit numbers can be formed from the digits 12345?
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
How many 5-digit numbers can be formed using the digits 0 1 2 3 4 5 6 7 and 8 if the first digit cannot be 0 or 1?
There are 7*94 = 45927 such numbers.
How many five digit numbers can be formed using the digits 02345 when repetition is allowed such that the number formed is divisible by 2 or 5 or both?
There are 2000 possible five digit numbers that can be formed from the digits 02345 that are divisible by 2 or 5 or both. To be divisible by 2, the last digit must be even, namely 0, 2 or 4 (in the digits allowed). To be divisible by 5, the last digit must be 0 or 5. Thus to be divisible by 2 or 5 or both, the last digit must be 0, 2, 4 or 5 (a choice of 4). Presuming that a 5 digit number must be at least 10000, then: For the first digit there is a choice of 4 digits (2345); for each of these there is a choice of 5 digits (02345) for the second, making a total so far of 4 x 5 numbers; for each of these choices for the first and second digits there is a choice of 5 digits (02345) for the third digit making the total so far (4 x 5) x 5 numbers; for each of these choices for the first three digits there is a choice of 5 digits (02345) for the fourth digit making the total so far (4 x 5 x 5) x 5 numbers; for each of these choices for the first four digits there is a choice of 4 digits (0245 - as discussed above) for the last digit, giving a total of (4 x 5 x 5 x 5) x 4 numbers. So the total number of five digit numbers so formed is: number = 4 x 5 x 5 x 5 x 4 = 2000.
How many 4 digits numbers can be formed using 0 1 2 3 4 5 6 7 8 9?
We figure out how many possibilities exist for each digit, and multiply them all together. Think of the digits as slots that need to be filled: _ _ _ _ The first digit can be anything but 0; 0345 is not a 4-digit number, for example. So there are 9 choices for the first digit. All the subsequent digits can be anything from 0-9; there are 10 choices for digits two, three, and four. So there are 9x10x10x10 = 9000 possible numbers with 4 digits.
How many three digit numbers can be formed using the digits 873 and 6 without repetition?
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
How many 3-digit numbers can be formed from the 5 digits 1 2 3 4 5 if no digit may be repeated in a number?
60 different numbers can be formed from the digits {1, 2, 3, 4, 5} is no repeats are allowed Any of the first digits can be chosen for the first digit, leaving 4 for the next and 3 for the final digit. Thus there are 5 × 4 × 3 = 60 different possible such permutations of 3 digits from the 5.
How many 7-digit phone numbers can be formed if the first digit cannot be 0 or 1 and any digit can be repeated?
The first digit can be formed in 8 ways (excluding 0 and 1). The rest of the 6 digits each can be filled in 10 ways. The total number of digits, therefore is 8 x 10^6.
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 3 digit numbers can be formed by 3 4 5 6 7 8 without repetition?
120. There are 6 digits. If we pick the digits in order, there are 6 possible digits for the first digit 5 remaining digits for the second digit 4 remaining digits for the third digit. 6*5*4 = 120.
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.
How many different positive four digit integers can be formed if the first digit must be even the last digit must be odd and all digits must be distinct?
There are 320 such numbers.
How many 10-digit numbers can be formed using the digits 1and 2?
You have 2 options for the first digit, 2 options for the second digit, etc. ... In total, that gives you 210 combinations.
How many 3 digits numbers can be formed from numbers 0-9?
The 3-digit numbers are all the counting numbers from 100 to 999.That's (the first 999 counting numbers) minus (the first 99 of them).There are 900 of them.
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 4 digit numbers can be formed from the digits 12345?
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
How many 5-digit numbers can be formed using the digits 0 1 2 3 4 5 6 7 and 8 if the first digit cannot be 0 or 1?
There are 7*94 = 45927 such numbers.
