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Easier to find the ones that aren't possible, then subtract that from 900 total 3-digit numbers.

Assuming we can use the same number more than once (e.g. 555)

Not possible:

100 140 150 160 170 180 190 => (7)

200 240 250 260 270 280 290 => (7)

300 340 350 360 370 380 390 => (7)

40* 440 450 460 470 480 490 => (16)

50* 540 550 560 570 580 590 => (16)

60* 640 650 660 670 680 690 => (16)

70* 740 750 760 770 780 790 => (16)

80* 840 850 860 870 880 890 => (16)

90* 940 950 960 970 980 990 => (16)

117 not possible

900 - 117 = 783 possible

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If A standard combination lock has 60 numbers on its face from 0 to 59 a combination consits of three numbers and numbers can be repeated what is the total number of possible combinations?

There are 60 possible numbers for the first number, A, in the combination (1,2,3,...,59,60).For each of outcome of A, there are 60 outcomes for the second number, B, giving a two digit combination 60x60=360 possibilities.For each of these 360 outcomes, there are 60 outcomes for the third number, C, making the number of possible combinations 360x60=21600.Or 60 possibilities x 60 possibilities x 60 possibilities = 21,600 possibilities.


What are all of the number combinations for the numbers 0 through 9?

If you can use the same number twice, such as 11 or 22, then there are 100 possible numbers (00, and 01-99). If you can not use the same number twice, the answer is 90.


How do you unlock 4 digit combination lock forgot number?

Your only solution - is to go through every possible combination from 0000 to 9999 until you find the correct number !


What is the equation to determine number of possible lock combinations?

In the simplest case, if A is the number of values a single element of a combination can have, and N is the number of elements in the combination, then the number of possible lock combinations is AN. For example: if you have a lock with a 4-digit numerical combination, any combination has 4 elements - the digits. Each digit can have 10 values - 0 through 9. So the total number of lock combinations is 104 or 10000. Another example: if you have the typical rotating combination padlock, a combination consists of three numbers, each of which can be 0-39. So the total number of combination shere is 403 or 64000. The calculation gets more complicated if the same number can't appear twice in the combination. In that case, there are A values for the first element, but only A-1 values for the second, A-2 values for the third, and so on. The formula in this case is A(A-1)(A-2)(...)(A-N+1) which can be written more concisely as A!/(A-N)!. For example: if you have the same rotating combination padlock as above, but know that the numbers in the combination are all different, there are 40*39*38 or 59280 possible combinations.


Is it possible for a rational number to be an integer but not a whole number?

No. Integers and whole numbers are the same.