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Q: How many permutations of 3 different digits are there chosen from the ten digits 0 to 9 inclusive?

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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.

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.

6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6

The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720

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.

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.

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.

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.

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.

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.

300

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.

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.

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.

8 digits will generate over 40,000 permutations.

If you wish to use a brute-force method, assuming the digits go from 0 to 9 there are 10,000 possible permutations.

There is nothing "in" digits. Digits are single characters representing the ten integers from 0 to 9 (inclusive).

900,000 of them. From 100,000 to 999,999 inclusive.900,000 of them. From 100,000 to 999,999 inclusive.900,000 of them. From 100,000 to 999,999 inclusive.900,000 of them. From 100,000 to 999,999 inclusive.

The sum of the number of digits in all the numbers between 31 and 400 inclusive is 1041.

Counting all integers from 1 to 238 inclusive, there are 606 digits between these two numbers.

Every number from 100 to 999 inclusive !

It is 180,001.

If you are talking about the combination function C(x,y) then your answer would be 210. However, try to be a little more specific. If you intend for permutations of 4 digit numbers (order matters), then you would have 10000 permutations for the digits, given that none of the digits are equal to each other.

This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.

The different digits have different values.

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