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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?
Mmcelarn
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 | -
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∙ 12y ago
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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
How many 4 number combinations are there in the numbers 2 to 6?
That's a supply of 5 digits . . . 2, 3, 4, 5, and 6.If the order of the four makes a difference, then there are (5 x 4 x 3 x 2) = 120 waysto select and arrange 4 out of the 5 digits.If the order doesn't matter, and you just want to know how many different groups of 4can be selected out of the 5 you have, then there are only 5 of those.
Is 0.044454647 irrational?
If those are all the digits, then it is rational.
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 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'.
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 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}.
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.
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 combinations are there for a 3-digit number if only 8 digits are usable?
-5
What are the duties of an ict officer?
solve the facts of the matter that contradict the conclusions of those that interpt the matters of life in order to live
How many 4 single digits combinations can add up to 10?
14
What are the first 120 digits of pie?
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647 Those are the first 120 digits of Pi
