Consider $S \in \mathbb{N}$ where $\mathbb{N}$ is the set of natural numbers. I want to obtain $S$ by summing $k$ natural numbers strictly positive. I want to find how many possibilities I have to obtain $S$ by summing $k$ numbers, considering that:  1) order does not matter, e.g. $2+1+3$ is the same as $3+1+2$  2) repetitions are allowed
114180
I want to find the sum of all combination of numbers whose sum is x, for e.g.  when x = 3 f(x) = countOf(111,12,21,3) = 4
109886
I have to give a combinatorial proof of  $$\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}.$$  I find it difficult to solve such problems. I'm not a brilliant person and never will be so I need to have algorithms to solve most problems. Usually I won't just "see" solutions unless I have a thorough enough understanding of a theory. This is also the case here. I know what the symbols mean, but I don't see what I should count and how to obtain the identity.  I think the right-hand side counts $n$-element subsets of a $(2n-1)$-element set with one element chosen to be "special". I think I should find a partition of the set of such subsets into $n$ parts so that the left-hand side is the sum of the cardinalities of the parts. Is there $n$ of anything on the right-hand side? Well, there are $n$ choices for the "special" elements, but only after I've chosen the $n$-element subset. Otherwise I can choose the "special element" in $2n-1$ ways.  This is a homework problem so I probably shouldn't ask for a full solution. But I would like to have some guidelines for approaching such problems, perhaps based on this example.
95699
I confirmed on  thread that there are as many as even natural numbers as there are natural numbers.   Question : Suppose I have selected a number $n \in \mathbb N$; what is the probability that $n$ is even?  My Thought : $\text{Probability} = \dfrac{\text{n(E)}}{\text{n(S)}}$ Here $\text{n(S)}$ is the set of all natural numbers i.e. $\mathbb N$, and $\text{n(E)}$ is set of all even natural numbers. Since it is proved that number of elements is the set $\mathbb N$ is exactly the same as the number of elements in the set of natural numbers (it’s very easy to put the set of natural numbers, $\Bbb N=\{0,1,2,3,\dots\}$, into one-to-one correspondence with the set $\text{E}=\{0,2,4,6,\dots\}$ of even natural numbers; the map $\Bbb N\to \text{E}:n\mapsto 2n$ is clearly a bijection.) ; Thus, Probability $= \boxed 1$ I know this is definitely wrong.Probability must be $0.5$. But where am I wrong? Can anyone explain ? Thanks!
117771
I'm looking for an alternative to the following (possibly standard) proof for the number of weak compositions:  The number of $k$-compositions of $n+k$ corresponds to a weak one by subtracting 1 from each "bin". Thus we have $\binom{n+k-1}{k-1}$.   While I kind of like this proof and always felt it made sense lately I've been left wanting something a little more convincing. Does anyone have a different proof of this result?
52877
So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a $S_n$ then $S_{n+1} = 4S_n$, but I didn't really succeed so here I am. Thanks in advance.
46126
QUESTION:   Prove that for all positive integers  $k\leq n$:  $\sum_{i=0}^{k}{n \choose i} (-1)^i = {n-1 \choose k}(-1)^k$  MY THOUGHTS:  INDUCTION : (assuming k even) - We assume:  ${n \choose 0} - {n \choose 1} + ... + {n \choose k-2} - {n \choose k-1} = {n-1 \choose k-1}$  We need to show:  ${n \choose 0} - {n \choose 1} + ...-{n \choose k-1} + {n \choose k} = {n-1 \choose k-1}$  My issue here is that we can say LHS equals ${n-1 \choose k-1} + {n \choose k}$, which is close, but not quite Pascal's Rule.   Another way: I thought looking to find some bijection that can be applied.   Let's assume n is even. Writing out the left and sum and moving all the odd (negative) terms to the RHS gives (note that the last term on the RHS is simply the original RHS:  ${n \choose 0} + {n \choose 2} + ... + {n \choose k} = {n \choose 1} + {n \choose 3} + ... + {n \choose k-1} + {n-1 \choose k}$    I am stuck here, but how I worked out in my mind was that:  $\bigg\{$set of even sized groups of  $\leq$ n people$\bigg\}$ -->  $\bigg\{$set of odd sized groups of  $\leq$ n-1 people$\bigg\}$ + ${n-1 \choose k}$  Then possibly, it would have something to do with adding or removing person n. Let me know your thoughts or suggestions.
27508
I've seen a question in my math book with this explanation above it: "If the product of n positive real numbers is 1, then the sum of these numbers must be more than n". I was wondering if this is correct, and if it is, could someone give an explanation or a proof for it?
5847
If $k$ is a positive natural number then $\phi(k)$ denotes the number of natural numbers less than $k$ which are prime to $k$. I have seen proofs that $n = \sum_{k|n} \phi(k)$ which basically partitions $\mathbb{Z}/n\mathbb{Z}$ into subsets of elements of order $k$ (of which there are $\phi(k)$-many) as $k$ ranges over divisors of $n$.  But everything we know about $\mathbb{Z}/n\mathbb{Z}$ comes from elementary number theory (division with remainder, bezout relations, divisibility), so the above relation should be provable without invoking the structure of the group $\mathbb{Z}/n\mathbb{Z}$. Does anyone have a nice, clear, proof which avoids $\mathbb{Z}/n\mathbb{Z}$?
38871
I'm having trouble deriving the following fact from the basic properties of $+_{\mathbb{N}}$ and the definition:  Definition. $\forall n, m \in \mathbb{N}: (n \geq m) \leftrightarrow (\exists a \in \mathbb{N}): n = m + a$. Fact. $\forall a, b \in \mathbb{N}: (a \geq b ~\land~ b \geq a) \rightarrow a = b$.  I've tried several different strategies, but none works. I think the fact that 0 isn't the successor of any natural number is the key both in the base case and the inductive step, but I keep going in circles trying to make use of that (and other basic facts about addition on $\mathbb{N}$). I would show you my notes, but there is really nothing of substance in them. I've made almost no progress since yesterday.
6479
For $n\in\mathbb{N}^*$ we note $s(n)$ the sum of the digits of $2^n$. For example $s(4)=7$ because $2^4=16$. Is it possible to find an integer $n$ such that $s(n)=s(n+1)$?
58502
Can we add an uncountable number of positive elements, and can this sum be finite?   I always have trouble understanding mathematical operations when dealing with an uncountable number of elements. Any help would be great.
49807
Let $m_1,m_2,\ldots,m_k$ be $k$ positive integers such that their reciprocals are in AP. Show that $k&lt;m_1+2$. Also find such a sequence.     Whatever way I tried, whichever formula I used, I could not eliminate the $m_k$ term, which I reckon would be the ideal scenario to find the solution. I am exhausted of any further ideas, please help. Thank you.
60821
Let $M$ be the set of natural numbers that can be written using only $0$ and $1$(in the decimal system). Prove that, for any natural number $k$ $\exists m \in M $ s.t   i) $m$ has exactly $k$ $1$'s, and ii) $m$ is divisible by $k$. (e.g, if $k=3$, then the number $101010 \in M$ is divisible by $3$)    Any hint or solution is welcome. If anyone have any hint in mind please give me step by step. Problem seems to be very interesting to me.
109872
For fixed positive integers $m$ and $n$, let $$S=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n.$$Show that $S = 0$ if $n &lt; m$ and $S = n!$ if $n = m$.  I know that I have to use PIE(principle of inclusion and exclusion) and a counting argument. Here's my work so far:  There are $m$ people, $\binom{m}{k}$ represents all the ways to pick $k$ people out of $m$ people. $(m-k)^n$ represents all the ways of giving the people that weren't chosen nn distinguishable objects. What I don't get is how to apply PIE, why the signs are alternating, and how does  $S = 0$ if $n &lt; m$ and $S = n!$ if $n = m$. Could anyone offer an answer by the counting argument I started? Thanks!
88366
Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help.
111505
Let  $n_1&lt;n_2&lt;n_3&lt;n_4&lt;n_5$  be positive integers such that $n_1+n_2+n_3+n_4+n_5=20$.Then what is the number of such distinct arrangements $(n_1,n_2,n_3,n_4,n_5)$?  My Approach :   I assumed   $n_1=t_0+1$  $n_2=n_1+t_1+1$  $n_3=n_2+t_2+1$  $n_4=n_3+t_3+1$  $n_5=n_4+t_4+1$  Where $t_0,t_1,t_2,t_3,t_4 \ge 0$  Now the sum becomes :  $5t_0+4t_1+3t_2+2t_3+t_4=5$  After this, I put the values of $t_i$s  $0,1,..$ and so on, and therefore found $7$ Solutions.  My question : Is there another way to solve this question, because as this question was asked in a competitive exam (JEE Advanced), This a very long solution.
112406
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.   From:   I see they mean that:  $$n-1 = \sum_{k=1}^{x} 2^{r_k} 3^{s_k}$$  By induction hypothesis. Then:  $$n = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} + 1$$  Then how to proceed by $n/2$? How can you say that:  $$\frac{n}{2} = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} $$  Where did the $1$ go? And how can you assume it works for $\frac{n}{2}$ (the hypothesis)?