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11

This can be seen from the fact that multinomial coefficients are integers : https://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients Then $\frac{(nk)!}{(n!)^k} = \binom{nk}{n,n,\dots,n}$, $\frac{(nk)!}{(k!)^n} = \binom{nk}{k,k,\dots,k}$ and $\frac{(nk)!}{(n!)(k!)} = (nk-n-k)!\binom{nk}{n,k,nk-n-k}$.


5

HINT: The order of the $2$ mathematics papers is significant. Multiply $10!$ by the number of ways in which you can arrange those items.


4

You are looking for an $(n,d,\lambda, \lambda)$ strongly regular graph. The normal labelling of the parameters are $(v,k, \lambda, \mu)$, so you are considering the special case where $\lambda = \mu$. There are lots of necessary conditions for the existence of a strongly regular graph. In particular, there are formulas to give the eigenvalues (there are ...


2

Break it into cases. If person $n-1$ is a man, use the induction hypothesis. If it's a woman, then the last person in line has to be a man, so... The base case is $n=2$, and I think you can probably prove it's true in that case.


2

$\frac{(kn)!}{n!^k}$ and $\frac{(kn)!}{k!^n}$ can both be recognized as multinomial coefficients which are integers. Since $n,k>1$ also $\frac{(kn)!}{n!k!(kn-n-k)!}$ is a multinomial coefficient.


2

Hint. One has $$ g'(x) = \left(\sum^\infty_{n=0}F_n \frac{x^n}{n!}\right)'=\sum^\infty_{n=1}n \times F_n \frac{x^{n-1}}{n!}=\sum^\infty_{n=1}F_n \frac{x^{n-1}}{(n-1)!}=\sum^\infty_{n=0}F_{n+1} \frac{x^n}{n!} $$ Can you do the same with $g''(x)$? Then use $$ F_{n+2}=F_{n+1}+F_n, $$ to get the announced identity.


2

For B. :$\binom {n+k} {k}=(n+k)!(n!!k!)^{-1}\in Z$ because it is the number of subsets of an $n+k$-element set that have exactly $k$ members each. So if $k>0$ then $k!$ divides the product of any $k$ consecutive positive integers, for if $k>0$ and $n\geq 0$ then $\binom {n+k} {k}= k!^{-1} \prod_{j=1}^k(n+j)).$ Therefore, for $n,i\geq 0 :$ ...


1

So we have three things we want to prove are divisible. By using modulo we may express this: a is divisible by b if: $$a \mod b = 0$$ Therefore, we can write the three equality's below and prove by lack of contradiction within algebra: A. $$(kn)! \mod (n!)^k = 0$$ Rewrite it as the large product operator: $$\prod_{i=1}^{kn} i \mod (\prod_{i=1}^{n} i)^k ...


1

A) and B) either both divide it or both don't by symmetry. We'll do A). To see that they both divide it, notice that the set $\{n,n+1,\ldots,2n-1\}$ can be put into bijection with $\{1,2,\ldots,n\}$ such that $x|f(x)$. Then we can break up $\{n,n+1,\ldots,2n,\ldots,nk\}$ into $k$ many sets that do this, and we are done, since now we can associate each ...


1

Since $S$ is isomorphic to a group of permutations on $26 - 9 = 17$ letters, its order is $17!$. Therefore it cannot contain an element of order $37$ or $38$, since neither of these numbers divide $17!$ (which is not divisible by any prime number greater than $17$). To show that it contains elements of order $36$ and $39$: factor these numbers into coprime ...


1

Note that $b=1$ is an impossibility as a cycle of length one that includes $1$ must have $\pi(1)=1$, not $\pi(1)=2$. Consider the case for when $b\geq 2$ Pick the $b-2$ remaining members of the cycle of length $b$ that includes $1$ followed by $2$ out of the $n-2$ possible remaining: Order the selections of the previous step within the cycle of ...


1

It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. We can write this way \begin{align*} [z^m](1+z)^n=\binom{n}{m}\tag{1} \end{align*} We obtain \begin{align*} \sum_{k=0}^{m}&\binom{n-p}{m-k}\binom{p}{k}\\ &=\sum_{k=0}^{\infty}[z^{m-k}](1+z)^{n-p}[u^k](1+u)^p\tag{2}\\ ...


1

Base case ($n=2$): man at back is right behind woman in front. So clearly a man is standing right behind a woman. Inductive Step: Assume true for $n=k$, so for any line $k$ people long there is a man standing right behind a woman. Then suppose you are faced with a line of $k+1$ people. Remove a person not at either end - the resulting line ($k$ people long) ...


1

$$\begin{align} \sum_{j=\color{red}0}^{n}{\sum_{k=0}^{j}{{n}\choose{k}}} &=\sum_{k=0}^n\sum_{j=k}^n\binom nk\\ &=\sum_{k=0}^n (n-k+1)\binom n{n-k}\\ &=\sum_{k=0}^n (j+1)\binom nj &&\text{putting }j=n-k\\ &=\sum_{k=0}^n j\binom nj+\sum_{k=0}^n \binom nj\\ &=n 2^{n-1}+2^n\\ \sum_{j=\color{red}1}^n\sum_{k=0}^n\binom ...


1

Since both the balls and the bags are distinguishable, what matters here is which balls are placed in which bag. Line up the bags in some order (for instance, by size or color). Choose which three of the $12$ balls go in the first bag, which three of the nine remaining balls go in the second bag, which three of the six remaining balls go in the third bag, ...



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