There are 5*4*3*2 = 120 such numbers.
106 or a million.
Just one. In a combination, the order of the digits does not matter.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
You can create several different numbers from the digits 1, 3, 4, and 6 in 1346. If you consider all possible combinations of these four digits, you can form a total of 24 unique four-digit numbers (4! = 24). Additionally, you can also create various three-digit, two-digit, and one-digit numbers, which would increase the total count significantly. However, counting all permutations of all possible lengths gives a comprehensive view of the total combinations.
This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.
90
7
106 or a million.
Just one. In a combination, the order of the digits does not matter.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
You can create several different numbers from the digits 1, 3, 4, and 6 in 1346. If you consider all possible combinations of these four digits, you can form a total of 24 unique four-digit numbers (4! = 24). Additionally, you can also create various three-digit, two-digit, and one-digit numbers, which would increase the total count significantly. However, counting all permutations of all possible lengths gives a comprehensive view of the total combinations.
This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.
Oh, dude, you're hitting me with the math questions, huh? Alright, so to find the number of 5-digit combinations from 1 to 20, you just do 20^5, which is like 3,200,000. So, yeah, there are 3,200,000 possible 5-digit combinations from 1 to 20.
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
You have 6 options for the first digit, 5 for the next, 4 for the next, and 3 for the next. Multiply all those numbers together.
you could make a probability tree if you could be bothered