20
19
There are twenty numbers between 19 through 39, including 19 and 39. So the problem is 20 times 20 times 20 times 20 times 20. 320,0000
63
Infinitely many. 1 + 1 + 1 + 19 .1 + .1 + .1 + 21.7 .01 + .01 +.01 + 21.97 And so on. And all these are with three of the numbers being the same. There are also combinations where some numbers are irrational -infinite, non recurring decimals. Then there are combinations with some of the numbers being negative.
20
19
There are twenty numbers between 19 through 39, including 19 and 39. So the problem is 20 times 20 times 20 times 20 times 20. 320,0000
63
Infinitely many. 1 + 1 + 1 + 19 .1 + .1 + .1 + 21.7 .01 + .01 +.01 + 21.97 And so on. And all these are with three of the numbers being the same. There are also combinations where some numbers are irrational -infinite, non recurring decimals. Then there are combinations with some of the numbers being negative.
Assuming you are using "combinations" in the mathematical sense where order doesn't matter (if order does matter it would be "permutations"), there are 22C5 = 22!/5!17! = 26,334 possible combinations of 5 numbers from 22. They start {1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, ... and end ... {16, 19, 20, 21, 22}, {17, 19, 20, 21, 22}, {18, 19, 20, 21, 22}; I'll leave the 26,328 combinations in the middle for you to list.
71
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)
Just one. When considering combinations, the order of the numbers makes no difference. So there is only one combination with those 7 numbers.
There are 10 whole numbers in 10 to 19.
two: 1 and 19
To calculate the number of combinations of 5 numbers possible from 1 to 20, we use the formula for combinations, which is nCr = n! / (r!(n-r)!). In this case, n = 20 and r = 5. Plugging these values into the formula, we get 20! / (5!(20-5)!) = 20! / (5!15!) = (20x19x18x17x16) / (5x4x3x2x1) = 15,504 possible combinations.