4

Note that your first bullet point is only true when $m \ge 4$, since otherwise all maximal subgroups of $G$ are abelian. If $m=3$ then there is nothing to prove so let's assume that $m \ge 4$. Now the derived subgroup $[G,G]$ has index $p^2$ in $G$ and it is equal to the intersection of any two maximal subgroups of $G$. So $[G,G]$ is abelian. Hence any ...


2

The $5$ indistinguishable marbles can be distributed in $6$ ways as follows: Give each person a single marble then you have $2$ marbles left. Now either $1$ of the person can get $2$ of these marbles in $3$ ways or you can give $2$ people $1$ marble each in $3$ ways so the marbles can be distributed in $6$ ways. Then to distribute the toys I can choose $1$ ...


2

First I'll count the right selections with ordering: The first has 100 options. (all are OK) The second has 95 options (1 gone, and 4 forbidden of the colour of the first) The third has 90 options (1 extra gone, 4 extra forbidden ones), and this pattern continues. (so $100-5(n-1)$ options for ball $n$) So in order we have $$100 \times 95 \times 90 \times \...


1

You have or $$number <number < number$$ or $$number = number < number$$ or $$number < number = number$$ So for each string you have 6 posibilities in 1. case and 3 posibilities in case 2. and 3. that is 12 posibilities. You have also posibilite that all are equal so you have 13 posibilites in total. What about $$number >number > ...


1

With $2$ "$=$", there is $1$ chance. $\\$ With $1$ "$=$", there is $3+3$ chances. ($3!/2$ different ordering for first $>$ then $=$ (Ex: $a>b=c$ (note that $a>b=c$ and $a>c=b$ is same)) and $3!/2$ different ordering for first $=$ then $>$ (Ex: $a=b>c$ (note that $a=b>c$ and $b=a>c$ is same)) ) $\\$ With $0$ "$=$", there is $6$ ...


1

Let $N$ be the number of 4 digit numbers that have at most 2 different digits occurring in them. There are $9$ numbers that have at most (so exactly $1$) digit(s). To get two digits exactly, we have two cases: we have a $0$ which does not occur in first place so $9$ other digits could be picked, and after that we can make $1 \times 2^3 - 1 = 7$ many ...


1

You have counted $1111$ several times, once for each 'other digit'


1

Silly overkill method: For any (positive) natural number $N$, set $$d_N=\sum_{abc=N}(a+b+c)\text{.}$$ Consider the Dirichlet series $$f(s)=\sum_{N}\frac{d_N}{N^s}\text{.}$$ Then $$\begin{split}f(s)&=\sum_{a,b,c}\frac{a+b+c}{a^sb^sc^s}\\ &=3\zeta(s-1)\zeta(s)^2\end{split}$$ so that $f(s)$ admits the Euler product $$\begin{align} \frac{f(s)}{3}&=\...


1

Consider $f(x)=2^5(x+\dfrac{2}{2})(x+\dfrac{3}{2})(x+\dfrac{5}{2})(x+\dfrac{7}{2})(x+\dfrac{11}{2})$ Your sum equals $3f(1)=3[2^5(1+\dfrac{2}{2})(1+\dfrac{3}{2})(1+\dfrac{5}{2})(1+\dfrac{7}{2})(1+\dfrac{11}{2})]=3(2+2)(2+3)(2+5)(2+7)(2+11)$


1

This answer preassumes that the order does not count so that also e.g. $A,M,T,H$ spells MATH. The number of possible outcomes if the letters come from $\{A,B,\dots,Z\}$ is $26P4\times10P4$. Similarly the number of possible outcomes if the letters come from $\{M,A,T,H\}$ is $4P4\times10P4$. So the searched probability is: $$\frac{4P4\times 10P4}{26P4\times ...


1

Presumably you mean $x^i(1-x)^{n-i}$ inside the sum. In that case, I think there is no nicer formula, except in special cases, such as $k=0$ or $k=n$ or $x=1/2$. Maple evaluates this in terms of a hypergeometric; but it is really just the definition of the series... $$ \sum _{i=k}^{n}{n\choose i}{x}^{i} \left( 1-x \right) ^{n-i} ={{n\choose k}\frac {{x}^{k}...


1

Under the assumption that $0\le a,b\le n$, we have $$a!(n-a)!=b!(n-b)!\iff{n!\over a!(n-a)!}={n!\over b!(n-b)!}\iff{n\choose a}={n\choose b}$$ To conclude that either $a=b$ or $a+b=n$, it is (more than) enough to show that ${n\choose k-1}\lt{n\choose k}$ for $1\le k\le\lfloor n/2\rfloor$. This inequality follows from $${n\choose k}={n\choose k-1}{n-(k-1)\...


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