Answer
$P(5, 3)=60$
The list of all arrangements:
$abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed\\
bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed\\
cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced\\
dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec
\\eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc$
Work Step by Step
If we want to choose $r$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $P(n, r)=\frac{n!}{(n-r)!}$ ways.
Hence,
$P(5, 3)=\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}=5\cdot4\cdot3=60$
The list of all arrangements:
$abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed\\
bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed\\
cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced\\
dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec
\\eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc$
