Question

List all the ordered arrangements of 5 objects $a,b,c,d$, and $e$choosing 3 at a time without repetition. What is$P(5,3)$?

Short answer

The list of ordered arrangements is:

$$\begin{gathered} abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed, \\ bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed, \\ cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced, \\ dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec, \\ eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc. \end{gathered}$$

The value of the permutation is 60.

Step-by-step solution

Step 1. Given information.

3 objects are chosen from 5 objects $a,b,c,d$, and $e$at a time without repetition.

The given permutation is $P(5,3)$.

Step 2. Determine the list of ordered arrangements.

The list of ordered arrangements is:

$$\begin{gathered} abc,abd,abe,acb,acd,ace,adb,adc,ade,aeb,aec,aed, \\ bac,bad,bae,bca,bcd,bce,bda,bdc,bde,bea,bec,bed, \\ cab,cad,cae,cba,cbd,cbe,cda,cdb,cde,cea,ceb,ced, \\ dab,dac,dae,dba,dbc,dbe,dca,dcb,dce,dea,deb,dec, \\ eab,eac,ead,eba,ebc,ebd,eca,ecb,ecd,eda,edb,edc. \end{gathered}$$

Step 3. Determine the value of permutation.

We know that,

$P(n,r)=\frac{n!}{(n-r)!}$

Using the above formula, we get

localid="1647083212967" $\begin{array}{rcl}P(5,3)& =& \frac{5!}{(5-3)!}\\ & =& \frac{5!}{2!}\\ & =& \frac{5\times 4\times 3\times 2!}{2!}\\ & =& 60\end{array}$