Question

How many $4$ digit numbers can be formed from digits $1,1,2,2,3,3,4,4,5,5$?

Answer 1

Step 1: Calculate the permutation

If $n$ is the total number of unique objects, and $r$ is the number of objects selected, then their permutation is,

${}^{n}P_r=\frac{n!}{\left(n-r\right)!}$

As all the digits are different, four digits can be formed in

$\begin{array}{l}{}^5{P}_4\end{array}$ ways$=\begin{array}{l}\frac{5!}{\left(5-4\right)!}\end{array}$

$=\frac{5!}{1!}$

$=120$ ways

Step 2: Calculate the combination with one repetition

If $n$ is total number of objects in the set, and $r$ is number of choosing objects from the set, then number of combinations,

${}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}$

With one repetition, the four digits can be formed in

$\begin{array}{l}{}^5{C}_1{\times }^4{C}_2\times \frac{4!}{2!}\end{array}$ ways$=\frac{5!}{1!\left(5-1\right)!}\times \frac{4!}{2!\left(4-2\right)!}\times \frac{4!}{2!}$

$=\frac{5!}{4!}\times \frac{4!}{2!\times 2!}\times \frac{4!}{2!}$

$=5!\times 3$

$=5\times 4\times 3\times 2\times 3$

$=360$ ways

Step 3: Calculate the combination with double repetition

With double repetition, the $4$ digit numbers can be formed in

$\begin{array}{l}{}^5{C}_2\times \frac{4!}{2!\times 2!}\end{array}=\frac{5!}{2!\left(5-2\right)!}\times \frac{4!}{2!\times 2!}$

$=\frac{5!}{2!\times 3!}\times \frac{4!}{2!\times 2!}$

$=\frac{5\times 4\times 3!}{2\times 3!}\times \frac{4\times 3\times 2}{2\times 2}$

$=5\times 4\times 3$

$=60$ ways

Step 4: Total number of $4$ digit numbers

Thus, the total number of $4$ digit numbers

$=120+360+60$

$=540$

Therefore, $540$ number can be formed from the digits $1,1,2,2,3,3,4,4,5,5$.