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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?
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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 possible 5 digit combInations are there with in ten digits?
In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.
How many 3-digit numbers can be formed using the digits 2 3 5 6 7 9 if repetition of digits is not allowed?
This is a basic permutation problem. You have 6 elements, of which you need to select 3, without replacement, and order does matter (123 is different from 321). Permutations are covered by the wikipedia page on permutations. In short, however the answer is P(6, 3) = 6!/(6-3)! = 6*5*4 = 120.
How many six digit numbers can be formed using the digits 0123456789 Example equals 123456 or 098765?
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
How many combinations are there if there are 8 numbers and you have a 16 number combination?
Assuming that "number" means digits and that it is permutations (rather than combinations) that are required, the answer is 816 which is 281.475 trillion (approx).
How many permutations of 3 different digits are there from the ten digits 0 to 9 inclusive?
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.
How many permutations of 3 different digits are there from the ten digits 0 to 9 inc?
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
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 possible 5 digit combInations are there with in ten digits?
In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.
How many 3-digit numbers can be formed from the 5 digits 1 2 3 4 5 if no digit may be repeated in a number?
60 different numbers can be formed from the digits {1, 2, 3, 4, 5} is no repeats are allowed Any of the first digits can be chosen for the first digit, leaving 4 for the next and 3 for the final digit. Thus there are 5 × 4 × 3 = 60 different possible such permutations of 3 digits from the 5.
How many license plate combinations from 3 digits followed by 3 letters?
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
How many 4 digit combinations can be made from 6 digits?
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
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 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 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 4 digit combinations can be made with 4 digits?
To calculate the number of 4-digit combinations that can be made with 4 digits, we can use the formula for permutations. Since there are 10 possible digits (0-9) for each of the 4 positions, the total number of combinations is 10^4, which equals 10,000. This is because each digit can be selected independently for each position, resulting in a total of 10 choices for each of the 4 positions.
