What steps will reproduce the problem?
1.copy the following text to latex-lab
documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\title{\LaTeX}
\date{}
\begin{document}
\maketitle
\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\title{\LaTeX}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item(1.9)
(a) If the domino is placed horizontally, we have $m$ choices for its horizontal position and $(n-1)$ for its vertical position.
Similarly, if the domino is placed vertically, we have $(m-1)$ choices for its horizontal position and $n$ for its vertical position.
Thus, totally in this case we have \[ m\times(n-1) + n\times(m-1) = 2mn - m - n \] tilings.
(b)We can solve this by first pick a position for one domino and then choose the position for the other domino.
Assume the first domino is vertically placed, we then eliminate the row that
contains the first domino and make the rest of them into one complete
$m\times(n-1)$ board left. By (a), we have $2m(n-1)$ options for placing the
second domino on this board. However, we need to add back the \[ (n - 2 - 2)+(m
- 2)- 2 = m + n - 8 \] possible options when the second domino interacts with
the eliminated column. Totally we have \[$2m(n-1)+ m + n - 8 = 2mn - m + n - 8
$. \]
By symmetry, we have \[2mn - m + n -8 + 2mn - n + m -8 = 4mn - 16 \]ways to put
exactly two dominoes on a board.
\item(2.3)
We can not pass through both $(8,7)$ and $(11,5)$ because $8<11$ and $7>5$.
So \[{13+9 \choose 13}-{8+7 \choose 8}{5+2 \choose 5}-{11+5 \choose 11}{2+4
\choose 2} = {22 \choose 13}-{15 \choose 8}{7 \choose 5}-{16 \choose 11}{6
\choose 2}\]
\item(2.4)
We start with the smallest number 1, it must be at the upper left corner of the
boxes. In other words, we must put the smallest them in one of the boxes in the
upper left region and the box should have two exposed sides. We say a box is
exposed if two or more of its sides are not shared with other boxes. We
eliminate a box after we put a number in it. Notice when we eliminate all the
boxes on a diagonal, it will give us one more exposed box. Otherwise, we have
the same number of exposed boxes as before.
\item(2.8) We have $(a_{1} + a_{2} + {\ldots} + a_{n})!$ ways of arrangement
for an alpha bet without repeating numbers. Since we have $ a_{i}! $ ways to
arrange the repeating letters, we totally have the multiple coefficient \[
{a_{1} + a_{2} + {\ldots} + a_{n} \choose a_{1} , a_{2} , {\ldots}, a_{n} } =
\frac{(a_{1} + a_{2} + {\ldots} + a_{n})!]{(a_{1} + a_{2} + {\ldots} +
a_{n})!\a_{1} ! a_{2} ! {\ldots} a_{n}!} \]
\item(2.9) \frac{2\times15!}{7!8!} because we have two types of endings with
the middle $8+7$ arrangement by the formula in 8.
\item(2.11)$ 366^{30} - 366!\frac{30!}{(366-30)!} = 366^{30} -
\frac{366!}/{30!336!} $
\end{enumerate}
\end{document}
2. I can only get the first compiled page and sometimes only numbers showed up.
3.
What is the expected output? What do you see instead?
What version of the product are you using? On what browser?
Please provide any additional information below.
Original issue reported on code.google.com by ccxc...@gmail.com on 22 Sep 2010 at 11:43
Original issue reported on code.google.com by
ccxc...@gmail.comon 22 Sep 2010 at 11:43