document.write( "Question 1165592: Given three red balls, four green balls and five blue balls, what is the probability that when you arrange those balls, no blue balls
\n" ); document.write( "are adjacent to each other?
\n" ); document.write( "a. 1/99 b. 4/99 c. 5/99 d. 7/99 \n" ); document.write( "

Algebra.Com's Answer #790080 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!
3 red balls, 4 green balls and 5 blue balls,
\n" ); document.write( "

\r\n" );
document.write( "We can arrange all the 7 non-blue balls in 7! ways.\r\n" );
document.write( "\r\n" );
document.write( "Below, think of the 7 X's as representing the 7 non-blue balls.\r\n" );
document.write( "Think of the 8 blanks as representing the 8 positions any 5 in \r\n" );
document.write( "which the 5 blue balls could be placed, between two non-blue\r\n" );
document.write( "balls, left of the first non-blue ball or right of the last\r\n" );
document.write( "non-blue ball.\r\n" );
document.write( "\r\n" );
document.write( "_X_X_X_X_X_X_X_\r\n" );
document.write( "\r\n" );
document.write( "For each of the 7! ways the non-blue balls could be placed,\r\n" );
document.write( "there are P(8,5) ways to place the 5 blue balls.\r\n" );
document.write( "\r\n" );
document.write( "That's (7∙6∙5∙4∙3∙2∙1)(8∙7∙6∙5∙4) = ways they could be placed\r\n" );
document.write( "out of 12! ways the balls could be placed with no restrictions:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "(7∙6∙5∙4∙3∙2∙1)(7∙6∙5∙4∙3)                     7\r\n" );
document.write( "--------------------------- = after canceling ---\r\n" );
document.write( "12∙11∙10∙9∙8∙7∙6∙5∙4∙3∙2∙1                     99\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( " \n" ); document.write( "

\n" );