I am trying to proof that for $n,m\in\mathbb{N}$ with $m\leq n$ the binomial coefficient $\binom{n}{m}=\frac{n!}{m!(n-m)!}$ is always a natural number. For I need to prove that $m!(n-m)!|n!$ for every $n,m\in\mathbb{N}$ with $m\leq n$.  Now, obviously the statement is true for $n=0$. Yet, once I assume that the statement is true for some $n$ I can't make proof that it holds for $n+1$. Here is where I got stuck: \begin{eqnarray} (n+1)! &amp;=&amp; (n+1)n!=(n+1-m)n! + n!m\\ &amp;=&amp; (n+1-m)!\frac{n!}{(n-m)!}+n!m\\ &amp;=&amp; (n+1-m)!\frac{km!(n-m)!}{(n-m)!}+n!m\\ &amp;=&amp; km!(n+1-m)!+n!m\\ \end{eqnarray} If anyone completed the proof by induction, I would be very thankful.  EDIT - I now have proven the statement this way using $\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}$:  $$\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}=k_0+k_1$$ with $k_0, k_1\in\mathbb{N}$, following our assumption. It follows that $$(n+1)!=(k_0+k_1)m!(n+1-m)!$$ thus proving the statement for $1\leq m\leq n$. The cases left are easy and I am omiting them here.  Still, I would like to see a prove by induction that utilizes the statement $(n+1)!=(n+1-m)n!+mn!$.  EDIT 2 - This is not a duplicate of , since, although the statement to proof is the same, what I am asking for is a proof by induction that uses a specific property of $(n+1)!$.  The proof I gave in my first edit was based on the second most up-voted answer given in the other post, but it is not what I am asking for.  As for the "best" answer given in the article this is supposedly a duplicate of: Neither that nor the answer it links to in another article constitute neither a proof of the statement, and much less a proof by induction.
12098
From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.  This is clearly an integer, however I was curious as to why the expression  $$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.  So far I figured:  $n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.
10141
Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
117849
How can one prove that for $n \equiv 0$ mod $2$  we have  $${n \choose 0} &lt; {n \choose 1} &lt;\ldots&lt; {n \choose n/2-1}&lt;{n \choose n/2}&gt;{n\choose n/2+1}&gt;\ldots&gt;{n\choose n-1}&gt;{n\choose n}\,?$$  Can I say that for fixed $n$, the binomial coefficients $n \choose k$ increase with $k$ for $k &lt; n/2$? If n is even (like in our case), then the central binomial coefficient $n \choose n/2$ is the largest one.   So $n \choose k+1$ is greater than, equal to, or less than $n \choose k$ according as $n-k$ is greater than, equal to, or less than $k+1$, that is according as $k$ is less than, equal to, or greater than $(n−1)/2$.  Is that correct?
28499
Prove that for every $n \ge m \ge 1 , \sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$    I've tried saying that the RHS represents the number of binary series with m "1" 's and n+1-m "0"'s, but I couldn't find out what k represents at the LHS. Thanks
53069
I need to proof this following equality :  $$\binom{m+n}{m}=\sum_{i=0}^m \left(\binom{m}{i}\binom{n}{i}\right)$$  This is what I did combinatoric proof:  Left : subset with $m$ members from $m+n$  Right : $\binom{m}{i} \rightarrow$ The number of subsets in set $m$ could also be written as $2^m$ AND $\binom{n}{i} \rightarrow$ the number of subsets with $m$ members from set $n$  In case $ n \leq m \implies \binom{n}{i}=2^n$  So this is what I came up with I still can't see why those expression are equal.  Any ideas? Thanks.
62005
4 . (a) Prove that   $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}.$$   Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.   I'm having a hard time trying to solve the problem above. I've done all of the previous exercises of the 2nd chapter with little difficulty, so far. I think I might be missing a trivial point somewhere.    The answer I got from the Answer Book, and is not very helpful either... :(      4. (a) Since $(1+x)^n(1+x)^m = (1+x)^{n+m}$  we have  $$\sum_{k=0}^n \binom{n}{k}x^k\cdot\sum_{j=0}^m \binom{m}{j}x^j\cdot=\sum_{l=0}^{n+m} \binom{n+m}{l}x^l$$  But the coefficient of $x^l$ on the left is clearly $$\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}.$$  One term of the sum occurring for each pair $k$, $j = l-k$.     I couldn't get the last part of the answer: why is it that the "coefficient of $x^l$ on the left is clearly $\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}$."?
115600
Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$  So far I have: Base case (n = 1) = $8^{1} | (4(1))!$  = $8 | 24$ which is true.  Induction Step:  $8^{n + 1} | (4(n + 1))!$  $8^{n + 1} | (4n + 4)!$   A bit confused as to how to close this proof out, also wanted to make sure my current progress is correct as well. Any help is appreciated.
33587
Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}  $  Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.   Good luck!  Here is what I got up to;  $\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!}    $
55194
Suppose that when $n=k$  $(k\geq4)$, we have that $k!&gt;2^k$.  Now, we have to prove that $(k+1)!\geq2^{k+1}$ when $n=(k+1)$ $(k\geq4)$.  $$(k+1)! = (k+1)k! &gt; (k+1)2^k  \text{ (since }k!&gt;2^k)$$  That implies $(k+1)!&gt;2^k2$, since $(k+1)&gt;2$ because of $k$ is greater than or equal to 4.  Therefore $(k+1)!&gt;2^{k+1}$  Finally  we may conclude that $n!&gt;2^n$ for all integers $n\geq4$    Solved, thank you!
111590
I am trying to prove that $\binom{2n}{n} &lt; \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) &lt; 4\sqrt{n(n+1)}$ which is false)
62356
Prove that $ \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-1)!k!}$.   Proof:  This is what I have tried:  $\frac{n!}{k!(n-k)!} \times (n-k)!  = \frac{n(n-1)\cdots(n-k+1)(n-k)!}{k(k-1)\cdots1(n-k)!}$  $\frac{n!}{k!} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)\cdots1}$  This is where my attempt ends.   Are my steps correct? If not how can I improve them?  New Questions:  How is it that n-k magically appears on the right side? I get that n-k and 3 times times 2 times 1 cancels out in the numerator and the denominator which leaves   $\frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)....1}$ which is equivalent to ${n \choose k}$ or \frac{n!}{(n-k)!k!}....is that correct?
19470
The left hand side has terms involving  $\binom{n}{m}= \dfrac{n!}{(n-m)!m!}$    $$1+\dfrac{1}{2}\binom{n}{1} +\frac{1}{3}\binom{n}{2}+...........+\frac{1}{n+1}\binom{n}{n} = \dfrac{2^{n+1}-1}{n+1}$$  I've done induction and proved $P(0)$ holds and also $P(1),P(2)$ holds (Just in case)  Now I've found $P(K+1)$ so the sum on the left is going to equal $\dfrac{2^{n+1}-1}{n+1}$ + $\dfrac{1}{n+2}$  and the right hand side is going to equal $\dfrac{2^{n+2}-1}{n+2}$  My problem is coming with the algebra though. I can't get those sides to equal. I can't get rid of the $n+1$ in the denominator.   Anyone have any ideas for me? They'd be much appreciated!  P.S. Someone gave me a hint and said that you can use the binomial theorem to solve this.
59439
Is there any way of determining if $\binom{n}{k} \equiv 0\pmod{n}$.  Note that I am aware of the case when $n =p$ a prime.  Other than that there does not seem to be any sort of pattern (I checked up to $n=50$).  Are there any known special cases where the problem becomes easier?  As a place to start I was thinking of using $e_p(n!)$ defined as:  $$e_p(n!) = \sum_{k=1}^{\infty}\left \lfloor\frac{n}{p^k}\right \rfloor$$  Which counts the exponent of $p$ in $n!$ (Legendre's theorem I believe?)  Then knowing the prime factorization of $n$ perhaps we can determine if these primes appear more times in the numerator of $\binom{n}{k}$ than the denominator.  Essentially I am looking to see if this method has any traction to it and what other types of research have been done on this problem (along with any proofs of results) before.  Thanks!
97505
Help finding a combinatorial proof of $k {n \choose k } = n {n - 1 \choose k -1}$  I have expanded it this far:  $$\frac{k \cdot n!}{k!(n-k)!} = \frac{n \cdot (n-1)!}{(k-1)!(n-k)!} $$  but then I am stuck from where to go from there.
65822
Please feel free to close this is necessary as I didn't see exactly this question (some variations that I tried but didn't seem to apply. Prove: $$\sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n}$$ I figured trying to muscle through it by expanding and then multiplying each term by variants of 1 such as $\frac{n^2}{n^2}$ and $\frac{n^2(n-1)^2}{n^2(n-1)^2}$ to get common denominators, but that was NOT the way it seemed as it got ugly pretty quickly.  Then I tried thinking up formulas of the binomial theorem that I could differentiate like other sums of binomial coefficient problems and i couldn't think of one.  Any help here would be great.
58589
I am trying to use mathematical induction to prove  $$(2n)!\ge2^n(n!)^2\quad\text{for }n\in\mathbb{N}$$  I am stuck at the $n=k+1$ point.
54052
I would appreciate if somebody could help me with the following problem  Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!} )$  $$\sum _{k=0}^n (n-2 k)^2 \binom{n}{k}=n\times  2^n$$