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How can you make a program of printing all possible combinations of any three digits?
Wiki User
∙ 14y ago
#include<iostream.h> void main() { int num[3],i,j,k; cout<<"enter any three numbers"; for(i=0;i<3;i++) cin>>num[i]; for(i=0;i<3;i++) for(j=0;j<3;j++) for(k=0;k<3;k++) if(i!=j && j!=k && i!=k) cout<<num[i]<<num[j]<<num[k]<<endl; }
Wiki User
∙ 14y ago
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What is the number of possible combinations using 7 digits?
128
How many possible 4 number combinations can you get using numbers 0-9?
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
How many number combinations possible from a 13 digit number?
There is only one possible combination of a 13 digit number created from 13 digits. In a combination, the order of the digits does not matter so that 123 is the same as 132 or 312 etc. If there are 13 different digits (characters) there is 1 combination of 13 digits 13 combinations of 1 or of 12 digits 78 combinations of 2 or of 11 digits and so on There are 213 - 1 = 8191 in all. If the characters are not all different it is necessary to have more information.
How many possible combinations of a 7-digit number?
If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.
If you wanted to know how many 5 digit combinations you could make using only 2 digits how would you solve this?
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
What is the number of possible combinations using 7 digits?
128
How many possible 4 number combinations can you get using numbers 0-9?
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
How many combinations of 5 are possible from 30 digits?
There are: 30C5 = 142,506
I have a Lock with 3 wheels and the digits 0 to 9 on it. what is the most combinations possible?
10 possible numbers on each wheel equals 10x10x10 or 1000 combinations possible.
Using 4 digits what is the total number of combinations for a 4 digit lock code?
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
How many different number combinations are in the world?
Since a number can have infinitely many digits, there are infinitely many possible combinations.
What is the number of possible 9 digit social security numbers if the digits can't be repeated?
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated
What combinations can you make with numbers 1472580?
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
How combination in a 9 digit number?
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
What combinations can you make with numbers 1 4 7 2 5 8 0?
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
Can you list all possible combinations there are from 0-9?
Perhapsf you specified the number of digits. e.g. 0-9 with two digits.
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.
