Question

If a $5$ digit number is created using the digits $1,2,3,3,5$. Then

• A. ${51}^{st}$ number from the beginning is $51332$
• B. ${51}^{st}$ number from the beginning is $51323$
• C. ${30}^{th}$ number from the end is $32153$
• D. ${30}^{th}$ number from the end is $32135$

Answer 1

Answer: (A), (D)

The correct options are
A ${51}^{st}$ number from the beginning is $51332$
D ${30}^{th}$ number from the end is $32135$

Available digits are : $1,2,3,3,5$
$\left(i\right)$Numbers starting with $1\left(1\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12\Rightarrow 12<51$ so move forward

Numbers starting with $2\left(2\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12+12=24\Rightarrow 24<51$ so move forward


Numbers starting with $3\left(3\_\_\_\_\right)$ are $=4!=24$ (as we have digit $3$ once)
Total numbers till now $=12+12+24=48\Rightarrow 48<51$ so move forward

Numbers starting with $5\left(5\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12+12+24+12=60\Rightarrow 60>51$
Now, numbers starting with $51$ $\left(51\_\_\_\right)$ are $\frac{3!}{2!}=3$
Total numbers till now $=12+12+24+3=51\Rightarrow$ that's what we are searching for.

$51233$ is at ${49}^{th}$ position
$51323$ is at ${50}^{th}$ position
$51332$ is at ${51}^{st}$ position.

$\left(i\right)$Numbers starting with $5\left(5\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12\Rightarrow 12<30$ so move forward.

Numbers starting with $3\left(3\_\_\_\_\right)$ are $=4!=24$ (as we have digit $3$ once)
Total numbers till now $=12+24=36\Rightarrow 36>30$ so move forward.
Now, numbers starting with $35$ $\left(35\_\_\_\right)$ are $3!=6$
Numbers starting with $33$ $\left(33\_\_\_\right)$ are $3!=6$
Numbers starting with $32$ $\left(32\_\_\_\right)$ are $3!=6$
Total numbers till now $=12+6+6+6=30\Rightarrow$ that's what we are searching for.

$32153$ is at ${29}^{th}$ position
$32135$ is at ${30}^{th}$ position

$\text{Alternate method}\left(ii\right)$

If any sequence has $n$ terms then , $k^{th}$ term from ending is $=(n-k+1{)}^{th}$ term from beginning
Since here, $n=\frac{5!}{2!}=60,k=30$
So, we need to find $n-k+1\Rightarrow 60-30+1={31}^{st}$ terms from beginning
Available digits are : $1,2,3,3,5$
Numbers starting with $1\left(1\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12\Rightarrow 12<31$ so move forward

Numbers starting with $2\left(2\_\_\_\_\right)$ are $=\frac{4!}{2!}=12$ (as we have digit $3$ twice)
Total numbers till now $=12+12=24\Rightarrow 24<31$ so move forward


Numbers starting with $3\left(3\_\_\_\_\right)$ are $=4!=24$ (as we have digit $3$ once)
Total numbers till now $=12+12+24=48\Rightarrow 48>31$
Now, numbers starting with $31$ $\left(31\_\_\_\right)$ are $3!=6$
Total numbers till now $=12+12+6=30<31$. So, move forward

numbers starting with $32$ $\left(32\_\_\_\right)$ are $3!=6$
Total numbers till now $=12+12+6+6=36>31$. First number starting with $\left(32\_\_\_\right)$ will be our required number that is $32135$