Wiki User
∙ 11y ago1296 or (6^4)
Wiki User
∙ 11y agoThere are 5,461,512 such combinations.
There are 1140 5 digit combinations from 1 to 20. 20 combination 3 computes that.
Allowing repetitions, there are 9 combinations. Without repeated digits, there is only one combination of 3 digits from 3.
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
100
There are 1140 five digit combinations between numbers 1 and 20.
There are 9C3 = 10*9*8/(3*2*1) = 120 of them.
Just 1.
7
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
Using the eight digits, 1 - 8 ,-- There are 40,320 eight-digit permutations.-- There is 1 eight-digit combination.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There are 167960 9 digits combinations between numbers 1 and 20.
Any 5 from 59 = 5006386
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
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.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.