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Will the number of permutations alwaysbe greater than the number of distinguisible permutations?
No, sometimes they will be equal (when all items being permutated are all different, eg all permutations of {1, 2, 3} are distinguishable).
What are the 120 combinations produced by 5 factorial?
5 factorial is just a number .... namely, 120 . If you want to see all of the 120 'permutations' of five objects, simply do this: -- List all of the 5-digit numbers from 01,234 to 43,210. -- Cross off every number on the list that has any digit greater than 4 in it. -- Cross off every number on the list that has any repeated digit in it. You'll have all of the 120 permutations of the objects 0, 1, 2, 3, and 4, without repetition.
How many 6 digit numbers are there that do not have the number 5 in?
The first digit can be any one of 8. (all except '0' and '5')The second, third, fourth, fifth, and sixth digit can be any one of 9. (all except '5')The total number of possible permutations is (8 x 9 x 9 x 9 x 9 x 9) = 472,392
Can you list for me all possible permutations for a 3 digit combination numbered 0 through 9 including repeats and triples?
No, there are 1000 of them and I have better things to do with my life!
What is the smallest 5 digit number with all different digit?
01234
How do you solve this program wap to print sum of a digit of an inputed number?
wap to print all the arnstrong no. between 100&1000
The number of permutations of n objects taken all together is?
The number of permutations of n objects taken all together is n!.
Will the number of permutations alwaysbe greater than the number of distinguisible permutations?
No, sometimes they will be equal (when all items being permutated are all different, eg all permutations of {1, 2, 3} are distinguishable).
What are the 120 combinations produced by 5 factorial?
5 factorial is just a number .... namely, 120 . If you want to see all of the 120 'permutations' of five objects, simply do this: -- List all of the 5-digit numbers from 01,234 to 43,210. -- Cross off every number on the list that has any digit greater than 4 in it. -- Cross off every number on the list that has any repeated digit in it. You'll have all of the 120 permutations of the objects 0, 1, 2, 3, and 4, without repetition.
How many 6 digit numbers are there that do not have the number 5 in?
The first digit can be any one of 8. (all except '0' and '5')The second, third, fourth, fifth, and sixth digit can be any one of 9. (all except '5')The total number of possible permutations is (8 x 9 x 9 x 9 x 9 x 9) = 472,392
How many 3 digit combinations can be made from 1 to 9?
All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell you. The number of combinations is 9C3 = 9*8*7/3*2*1 = 84
What is the result of symbol number 2085004?
There is no result: all you have is a 7-digit number.There is no result: all you have is a 7-digit number.There is no result: all you have is a 7-digit number.There is no result: all you have is a 7-digit number.
How many 6 digit numbers can be formed with 4 different digits?
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.
What are the possible combinations in a 4 digit password?
Passwords are technically permutations, not combinations. There are 104 = 10000 of them and I regret that I do not have the time to list them. They are all the numbers from 0000 to 9999.
Can you list for me all possible permutations for a 3 digit combination numbered 0 through 9 including repeats and triples?
No, there are 1000 of them and I have better things to do with my life!
What is the smallest 5 digit number with all different digit?
01234
What are all the 4 digit combinations with 0-9?
Method 1: four digit number start after 999 and end at 9999so 4 digit no. = 9999-999 = 9000method 2: - - - - there are four places in 4 digit no. In first place only 1-9 can come , for rest 0-10so total no. = 9*10*10*10=9000* * * * *Wrong!Those are the number of PERMUTATIONS, not combinations. In a combination, the order of digits does not matter. So 1234 is the same as 1342 or 4213 etc.
