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The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE!

If the 30 numbers are the first 30 counting numbers then there are

126 combinations of five 1-digit numbers,

1764 combinations of three 1-digit numbers and one 2-digit number, and

1710 combinations of one 1-digit number and two 2-digit numbers.

That makes a total of 3600 5-digit combinations.

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Q: How many 5 digit combos in 30 numbers?

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There are 30 such numbers.

Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.

The first 3 digit integer being a positive multiple of 30 is 120. The final 3 digit integer being a positive multiple of 30 is 990. 990 - 120 = 870. 870 ÷ 30 = 29 But, as 29 is the difference between the two limits and the limits themselves are included then there are 29 + 1 = 30 such numbers.

30.The first digit can be one of three digits {3, 6, 9} corresponding to the last digit being {1, 2, 3}, and for each of those three digits, the middle digit can be one of ten digits {0 - 9}, making 3 x 10 = 30 such numbers.It is assumed that a 3 digit number is a number in the range 100-999, excluding numbers starting with a leading zero, eg 090 is not considered a 3 digit number (though it would be a valid 3-digit number for a combination lock with 3 digits).

There is no limitation given to the number of digits in the number, thus: For 3 digit numbers: to be less than 500 the first digit can be only one of {1, 2, 3, 4} leading to 4 × 5 × 4 = 80 numbers For 2 digit numbers: they are all less than 500, and there are 6 × 5 = 30 numbers For 1 digit numbers: they are all less than 500, and there are 6 numbers Thus there are in total 80 + 30 + 6 = 116 numbers that can be made form the digits {1, 2, 3, 4, 5 ,6} without repetitions that are less than 500.

Related questions

There are 30 such numbers.

29 of them.

There are 300.

Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.

30

30

30

30 of them.

If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.

300600900are some

13

The digit zeroes appear in the numbers 10, 20, 30, 40 and 50, making a total of 5 digit zeroes.