The number of 3-digit numbers with no repeated digits is simply 10x9x8 = 720, if you allow, for example, 012 as a 3-digit number.
There are 10 digits, any of which might be the first digit. The second digit can be any digit except the digit that was used for the first digit, leaving 9 possibilities. The third digit then has 8 possibilities, since it can't be the same as the first or second digit.
The actual number of possible area codes will be lower, because there are additional restrictions on the number combinations for a valid area code. For example, in North America (USA, Canada, etc.), the first digit of an area code cannot be 0 or 1 and the middle digit cannot be 9.
No, it is irrational because the decimal goes on without repeating.
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
A repeating decimal is a rational number. Its value is(the repeating set of digits)/(as many 9s as there are digits above).
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Yes the square root of 150 is 12.247448713915890490986420373529This is irrational because the answer is a number that the decimal goes on forever without repeating.
4321
987654321
10234567
9,876,543
The least possible integer is -98765432. The least possible positive integer is 10234567.
There are twelve possible solutions using the rule you stated.
The smallest possible 9 digit number (assuming negatives are not allowed) is 102,345,678 in which 5 is in the thousands' place.
12345 but if you can use a decimal, then 0.1234 would be the smallest possible number.
It would contain of all the number from 1000 to 9998 leaving 9 numbers.
There are only five combinations: 1234, 1235, 1245, 1345 and 2345.
Assuming whole numbers and a leading zero does not count: 10,234,567
9876543