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How many combinations order matters can you get from 4 digits where two of those digits are the same eg 8533?
I completed the calculation by brute force, but I would like to know the formula behind it:
The first column is all combinations with 4 different numbers {1,2,3,4}.
The second column is all combinations with 3 different numbers in 4 places where two of the numbers are the same {1,2,3,3}
1234 | 12331243 | -1324 | 13231342 | 13321423 | -1432 | -2134 | 21332143 | -2314 | 23132341 | 23312413 | -2431 | -3124 | 31233142 | 31323214 | 32133241 | 32313412 | 33123421 | 33214123 | -4132 | -4213 | -4231 | -4312 | -4321 | -
What else can I help you with?
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 combinations of 9 digits are possible if no two adjacent digits are the same?
The first can be any one of 10 digits. For each of those . . .The second can be any one of the nine digits not the same as the first. For each of those . . .The third can be any one of the nine digits not the same as the second. For each of those . . .The fourth can be any one of the nine digits not the same as the third. For each of those . . .The fifth can be any one of the nine digits not the same as the fourth. For each of those . . .The sixth can be any one of the nine digits not the same as the fifth. For each of those . . .The seventh can be any one of the nine digits not the same as the sixth. For each of those . . .The eighth can be any one of the nine digits not the same as the seventh. For each of those . . .The ninth can be any one of the nine digits not the same as the eighth.The total number of possibilities is: (10 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9) = 430,467,210 .If matching adjacent digits were allowed there would be 1,000,000,000 possibilities.So the restriction has eliminated 56.95% of them.
What so special about this number 854917632?
Only for those who speak English: the digits of the number are in alphabetic order.
How many 4 single digits combinations can add up to 10?
14
What combination of numbers can you get from 3789?
Well honey, there are quite a few combinations you can whip up with those digits. You can get numbers like 3789, 7983, 9837, and so on. Just mix 'em up like a cocktail and see what you come up with!
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 combinations for numbers 1248?
To calculate the number of combinations for the numbers 1248, we need to consider all possible arrangements of the four digits. Since all the digits are unique, there are 4 factorial (4!) ways to arrange them. This equals 4 x 3 x 2 x 1 = 24 combinations.
How many 3 number combinations are there in the numbers 0 5 7?
There are 6 different ways (permutations) to write those 3 digits. If the sequence (order) doesn't matter, then there's only one 'combination'.
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.
What are the possible different combinations of numbers in four digits?
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
How many one two or three digit combinations can you get using 1 2 and 3 in any one two or three configuration?
This question could would be best answered if you have a basic understanding of combinatorics (the study of combinations and permutations and probability) however, for something with as few possibilities as this, it is often easier to just make a list. Start with three digit combinations, then two digit combos, then one digit. Remember also that combinations are order independent ( ABC is the same thing as CBA) 3 digits 3 2 1 2 digits 3 2 3 1 2 1 1 digits 1 2 3 count those up and voila.
How many combinations of 9 digits are possible if no two adjacent digits are the same?
The first can be any one of 10 digits. For each of those . . .The second can be any one of the nine digits not the same as the first. For each of those . . .The third can be any one of the nine digits not the same as the second. For each of those . . .The fourth can be any one of the nine digits not the same as the third. For each of those . . .The fifth can be any one of the nine digits not the same as the fourth. For each of those . . .The sixth can be any one of the nine digits not the same as the fifth. For each of those . . .The seventh can be any one of the nine digits not the same as the sixth. For each of those . . .The eighth can be any one of the nine digits not the same as the seventh. For each of those . . .The ninth can be any one of the nine digits not the same as the eighth.The total number of possibilities is: (10 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9) = 430,467,210 .If matching adjacent digits were allowed there would be 1,000,000,000 possibilities.So the restriction has eliminated 56.95% of them.
How many combinations of numbers can be made of 3 numbers if the numbers can be repeated?
10 to the third assuming zero is included and can lead, otherwise 9 to the third, ie 729. This all assumes that you are talking about 3-digit numbers. * * * * * No. That may be the number of permutations but those are different from combinations. In a combiation, the order of the digits does not matter so that 123 is the same as 132 or 213 etc. With repetition, there are 210 COMBINATIONS, including one that is {0,0,0}.
Can you make any numbers with those ten digits?
depens if the number can be used in any order
What so special about this number 854917632?
Only for those who speak English: the digits of the number are in alphabetic order.
How many numbers can be formed using digits 1245?
Well honey, you've got 4 digits there, so you can form 4! (4 factorial) which is 24 numbers. That's right, you can make 24 different combinations with those digits. Math can be fun when you've got some sassy numbers to play with!
How many combinations are there for a 3-digit number if only 8 digits are usable?
-5
