Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 23


The ordered arrangements: $abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed$ $bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed$ $cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced$ $dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec$ $eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc$ $P(5,3)=60$

Work Step by Step

The ordered arrangements: $abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed$ $bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed$ $cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced$ $dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec$ $eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc$ We know that $P(n,r)=n(n-1)(n-2)...(n-k+1).$ Also $P(n,0)=1$ by convention. Hence, $P(5,3)=5\cdot4\cdot3=60$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.