The first digit can be from 0 through 9, the seconds digit can be from 0 through 9 and the third digit cane be 0 through 9, then we have thr total number of tags as
(10 x 10 x 10) = 1000 tags, if we assume that 000 is a valid tag
If however, 000 is not a valid tag, then you can make 1000 -1 = 999 tags
4 and 7
Assuming a 3-digit password can begin with a zero, the answer is 720.
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
Number of possible groups of 3 letters = 26 x 25 x 24 = 15,600. For each of these . . .Number of possible groups of 3 digits = 9 x 9 x 8 = 648 .Total number of possible distinct plates = 15,600 x 648 = 10,108,800
If you can repeat a digit, there are 27. If you can't repeat a digit, there are only 6.
6 of them.
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
1000: from 000 to 9991000: from 000 to 9991000: from 000 to 9991000: from 000 to 999
Your answer is, depending on the order of your subtraction, either positive or negative 198
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
All numbers are made with digits and there is no limit on how big they can be. So there is no greatest number.
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
36 of them.
12
Six. 321 312 231 213 123 132
Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits. There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively. So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120 Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360. the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720. The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720. So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.
45