# Tag Info

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)\...

Only top voted, non community-wiki answers of a minimum length are eligible