Answer
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,
$P(5,3)=5\cdot4\cdot3=60$.
