document.write( "Question 253885: Could you please help me with the following question?\r
\n" ); document.write( "\n" ); document.write( "A doctor visits 5 people by bus once a day. He can use either of two buses, but once he starts out he visits all 5 people in the .
\n" ); document.write( "same order before returning home. How many possibilities are there?
\n" ); document.write( " \r
\n" ); document.write( "\n" ); document.write( "Thank you
\n" ); document.write( "

Algebra.Com's Answer #186225 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Since order matters, he has 5 choices for the first patient, 4 for the second (since he can't see the same person twice), 3 for the third, 2 for the fourth, and one for the fifth.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Multiply these choices out to get 5*4*3*2*1=120\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So there are 120 different possibilities. \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Alternatively, you can use the permutation formula \"P%28n%2Cr%29=n%21%2F%28n-r%29%21\" with n=5 and r=5 to get \"P%285%2C5%29=5%21%2F%285-5%29%21=5%21%2F0%21=%285%2A4%2A3%2A2%2A1%29%2F1=120%2F1=120\" and you'll get the same answer
\n" ); document.write( "
\n" );