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 Apr26 answered How to deduce the formula “distribution” in groups? What is the difference between “distribution” & “arrangement”? Mar16 comment Stirling number of the second kind and combinations Actually i wanted to create a function which maps element from the (n-r)-sized subset to the set of r-partitions. If you have a function like this which is either onto or one-one then you are done with your proof. Mar16 answered Stirling number of the second kind and combinations Feb6 comment Loss functions for regression You have used x and x' both , so i was little confused. Feb6 comment Loss functions for regression Thanks for your answer. However, i have a doubt. I would be grateful if you could clarify that. Here, y(x) is the predicted value for x . And in the training data expected input-output pairs are give. Both x and y(x) are real-valued. So, i am not sure why we need x' in the computation? Feb5 asked Loss functions for regression Feb1 comment Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T. you can't use the recursion for F directly. When you have the recursion you can simply unwind it in this way F(n) = n-2 + F(2) = n-1 to obtain the answer, without using induction. On the other hand by proving the fact by induction you obtain F(n) = n-1. Also note F(n) is defined for n=1. Feb1 answered Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T. Jan26 answered Histogram of duplication in n choose k Jan8 revised Representing “not” in lambda calculus added 357 characters in body Jan7 comment Representing “not” in lambda calculus what i tried to mean is f (or P, i just edited) does not have any more lambda inside it. Jan7 revised Representing “not” in lambda calculus edited body Jan7 asked Representing “not” in lambda calculus Jan6 comment Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s yes, it is. Thanks for pointing. Jan6 revised Use the PIE to prove an identity added 2 characters in body Jan6 revised Use the PIE to prove an identity added 133 characters in body Jan6 asked Use the PIE to prove an identity Jan6 revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s added 240 characters in body Jan6 revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s Slight correction in the formula Jan6 revised Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s added 12 characters in body