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In Number of non-decreasing functions?,

I know that by adding a strictly increasing function(which in this case is n-1) to a non-decreasing function, we are creating another strictly increasing function. In this case range comes out to be {1..29}.

However, we can create an increasing function by adding any increasing function to a non-decreasing function. Why are we taking only n-1 in this case ?

For example, if we add 2n(which is also a strictly increasing function) to the non-decreasing function, the result is also a strictly increasing function. But in this case, range varies from {3...40}.So, the answer comes out to be $38 \choose 10$ which is not equal to $29 \choose 10$ ?

Where am I going wrong ?

Random Walker
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  • The point is to come up with something that results in a bijection, so that each $10$-set results in exactly one non-decreasing function. Other choices either will end up with $10$-sets that don't correspond to a non-decreasing function or with multiple $10$-sets that correspond to the same non-decreasing function. In your example there will be $10$-sets that don't correspond to any non-decreasing function. – Robert Shore Dec 21 '20 at 19:56
  • @RobertShore Okay, for range [31...40] , we cannot find a suitable non-decresing function f(n) – Random Walker Dec 21 '20 at 20:38

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