41
There are 18C4 = 18!/[18-4)!4!] = 18*17*16*15/(4*3*2*1) = 3060 combinations.
To answer this question you need to start with the "... combinations you wrote for 12, 18 and 21". Notice that the numbers in the second set are each 10 times the numbers in the first set. Combinations for numbers in the second set can be found by adding "*10" or "*5*2" to the combinations for the first set. You could also say: 12 = 2*2*3; 120 = 2*2*3*5*2 = 4*15*2, and many other variations on this theme.
The set of even numbers
It depends how many numbers are in your combinations: 0 or 24 there is 1 1 or 23 there are 24 2 or 22 there are 276 3 or 21 there are 2,024 4 or 20 there are 10,626 5 or 19 there are 42,504 6 or 18 there are 134,596 7 or 17 there are 346,104 8 or 16 there are 735,471 9 or 15 there are 1,307,504 10 or 14 there are 1,961,256 11 or 13 there are 2,496,144 12 there are 2,704,156 If by combinations you do not mean the mathematical selection of a subset of items from a set of items (where the order doesn't matter) but how many ways are there of arranging 24 numbers then there are: 24! = 620,448,401,733,239,439,360,000 ways.
121
there are many combinations of numbers that add up to 84. Some examples are 41+43, 35+49, 29+55, 24+61, and 18+66.
233 = 18 combinations.
18. 18 x 3 = 54. However, since the question asks for even numbers, many other combinations would qualify (e.g. 40 + 8 + 6, 32 + 12 = 10).
12 and 18
The first number has 19 possibilities. The second has 18, and the third only 17. 19x18x17=5814 possible lock combinations. (If you could repeat, it would be 19x19x19=6859)
The mean is the average of a set of numbers. To find the average, you must add all the numbers in the set and divide that number by however many numbers are present in the set. For this example, you need to add 32 and 4 and then divide that answer by two, which is 18 in this case.
Three possible combinations: 17+1, 13+5 and 7+11.