2,704,156
100
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
Assuming that the only permitted operation is addition, then there are 4 combinations.
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.
There are 12 different combinations of 3 positive odd numbers that add up to 21. Namely: (There are many permutations of these combinations.) 1,1,9 1,3,17 1,5,15 1,7,13 1,9,11 3,3,15 3,5,13 3,7,11 3,9,9 5,5,11 5,7,9 7,7,7
There are millions of combinations of numbers with a difference of 9. 9 and 0, 10 and 1, 11 and 2, 12 and 3, and so on. The difference is the answer that you get when you subtract two numbers, so any numbers that subtract to an answer of 9 will work.
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
You would get 4!/2! = 12 combinations.
30
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
26 = 64 combinations, including the null combination - which contains no numbers.
Assuming that the only permitted operation is addition, then there are 4 combinations.
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
There are 14C5 = 14*13*12*11*10/(5*4*3*2*1) = 2002
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.
11
This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.
There are 12 different combinations of 3 positive odd numbers that add up to 21. Namely: (There are many permutations of these combinations.) 1,1,9 1,3,17 1,5,15 1,7,13 1,9,11 3,3,15 3,5,13 3,7,11 3,9,9 5,5,11 5,7,9 7,7,7