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How many permutations of 3 different digits are there from the ten digits 0 to 9 inclusive?
Wiki User
∙ 13y ago
ABC
A = 1,2,3,4,5,6,7,8,9; 9 possibilities
B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities
C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities
So there are 9*10*10 = 900 numbers with 3 digits.
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∙ 13y ago
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How many 6 digit numbers can be formed with 4 different digits?
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.
How many positive integers have exactly three digits?
Every number from 100 to 999 inclusive !
How many different permutations are in the word ASSESSMENT?
There are 75600 permutations.
How many 3 digit numbers can be formed using the digits 2 3 5 6 7 9 if repetition of digits is not allowed?
120 There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4 = 120 of them.
How many digits are on a passport?
Different countries have different number of digits. You need to be more specific.
How many permutations of 3 different digits are there chosen from the ten digits 0 to 9 inclusive?
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
How many permutations of 3 different digits are there from the ten digits 0 to 9 inc?
There are 720 permutations of 3 different digits from the ten digits 0 to 9, including leading zeros.
How many different numbers can be formed with the digits 345?
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.
How many permutations of two different digits can be obtained from a set of four different digits?
6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6
How many number can be created with 15679?
It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.
How many 4 digit combinations can be made from 6 digits?
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.
How many 4 digit combinations can be made with 4 digits?
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
How many different two digit numbers can you form using the digits 1 2 5 7 8 and 9 without repetition?
36 two digit numbers can be formed...(:From Rafaelrz: The question can be stated as;how many permutations of two different digits can beobtained from a set of six different digits ?Answer:nPr equals n!/(n-r) ...... for n = 6, r = 26P2 equals 6!/(6-2)! equals 30 Permutations.
How many digits are there from 100-200?
300
How many 4 number combinations using digits 1357 one and only once?
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
How many numbers used for 10000 permutation?
8 digits will generate over 40,000 permutations.
How many 6 digit numbers can be formed with 4 different digits?
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.
