用C# 实现启动应用程序?
怎么用C#   实现启动应用程序,如word,要所有的应用程序通用?
谢谢

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MSDN 上的例程

using System;
using System.Diagnostics;
using System.ComponentModel;

namespace MyProcessSample
{
/// <summary>
/// Shell for the sample.
/// </summary>
class MyProcess
{

/// <summary>
/// Opens the Internet Explorer application.
/// </summary>
void OpenApplication(string myFavoritesPath)
{
// Start Internet Explorer. Defaults to the home page.
Process.Start( "IExplore.exe ");

// Display the contents of the favorites folder in the browser.
Process.Start(myFavoritesPath);

}

/// <summary>
/// Opens urls and .html documents using Internet Explorer.
/// </summary>
void OpenWithArguments()
{
// url 's are not considered documents. They can only be opened
// by passing them as arguments.
Process.Start( "IExplore.exe ", "www.northwindtraders.com ");

// Start a Web page using a browser associated with .html and .asp files.
Process.Start( "IExplore.exe ", "C:\\myPath\\myFile.htm ");
Process.Start( "IExplore.exe ", "C:\\myPath\\myFile.asp ");
}

/// <summary>
/// Uses the ProcessStartInfo class to start new processes, both in a minimized
/// mode.
/// </summary>
void OpenWithStartInfo()
{

ProcessStartInfo startInfo = new ProcessStartInfo( "IExplore.exe ");
startInfo.WindowStyle = ProcessWindowStyle.Minimized;

Process.Start(startInfo);

startInfo.Arguments = "www.northwindtraders.com ";

Process.Start(startInfo);

}

static void Main()
{
// Get the path that stores favorite links.
string myFavoritesPath =
Environment.GetFolderPath(Environment.SpecialFolder.Favorites);

MyProcess myProcess = new MyProcess();

myProcess.OpenApplication(myFavoritesPath);
myProcess.OpenWithArguments();
myProcess.OpenWithStartInfo();

}
}
}


------解决方案--------------------
http://blog.csdn.net/chengking/archive/2005/10/07/496752.aspx
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Process.Start(filename);
Help.ShowHelp(filename);
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System.Diagnostics.Process.Start(@ "C:\file.exe ");
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using System.Diagnostics;

Process MyProcess = new Process();
MyProcess.StartInfo.FileName = @ "D:\Program Files\Tencent\QQ\QQ.exe ";
MyProcess.StartInfo.Verb = "Open ";
MyProcess.StartInfo.CreateNoWindow = true;
MyProcess.Start();