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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.

Q: How many 5 numbers combinations can you make from the numbers 1 to 22 and what are they?

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There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.

If each number is different from all of the other numbers and the combinations are considered distinct unless they contain the same numbers in the same order, the answer is 25 X 24 X 23 X 22 X 21 = 6,375,600. If the combinations are distinct only if they do not contain the same numbers, irrespective of order, then the answer is (25 X 24 X 23 X 22 X 21)(5 x 4 x 3 x 2) = 53,130.

If the numbers are all different, then 25C5 = 25*24*23*22*21/(5*4*3*2*1) = 53,130.

1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations

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.

Related questions

There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.

22

11

If each number is different from all of the other numbers and the combinations are considered distinct unless they contain the same numbers in the same order, the answer is 25 X 24 X 23 X 22 X 21 = 6,375,600. If the combinations are distinct only if they do not contain the same numbers, irrespective of order, then the answer is (25 X 24 X 23 X 22 X 21)(5 x 4 x 3 x 2) = 53,130.

If the numbers are all different, then 25C5 = 25*24*23*22*21/(5*4*3*2*1) = 53,130.

1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations

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.

22

umm i think there would be 3 combonations. correct this if you think this is wrong. 00, 11, 22,

If there are no restrictions on the 'combinations' then there are ten choices for the first digit and ten for the second: 10 x 10. (This implies possibilities such as 22 and 77.) If the digits must be different in each combination then the number of combinations taking two at a time from ten is C(10,2) = 10!/( 2! ( 10 - 2 )! ) = 45.

It looks like you are asking how many combinations of 6 numbers are there in the 28 numbers 1 through 28. This is known as the number of combinations of 28 things taken 6 at a time. The answer is 28!/(6!22!) (n! means n factorial, which is the product of all the integers from 1 to n). I get 376,740 if I haven't made an error in arithmetic.

Answer : In total there are 22 numbers.