Possible combinations of numbers?

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To determine the number of combinations of numbers picked by 10 people from a range of 1 to 120 without repetition, the calculation involves decreasing choices for each subsequent picker. The first person has 120 options, the second has 119, the third has 118, and this pattern continues until the tenth person, who has 111 choices. The total number of unique combinations can be calculated as 120 × 119 × 118 × 117 × 116 × 115 × 114 × 113 × 112 × 111. This approach ensures that no two numbers are the same among the selections.
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Homework Statement


Assuming 10 people pick a number between 1-120 and no two numbers can be the same, how many combinations of numbers can their be?

Homework Equations


C=N^s

The Attempt at a Solution


So I know the total possible for 120 would be just 120^10, but if I want to solve for the numbers never repeating in a sequence I'm not sure how.
 
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The first guy picks one number from 1 to 120. So he has 120 choices. But the second guy has 119 choices because one number is taken by the first guy. Similarly the third guy has 118 choices. And so on.
 

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