Question

All the five-digit numbers in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The ${{97}^{th}}$ number does not contain the digit?

Answer 1

The series of five-digit numbers contains a unique property where each digit succeeds its predecessor. 12345 is the first term of this series as in this number all the digits exceed its predecessor. We will start counting all the five-digit numbers starting from 1 at first and then move on to five-digit numbers starting with 2 and so on until we land on the ${{97}^{th}}$ number of the series. This will give us the required answer.

Complete step by step answer:
The first set of numbers that can be formed are of “1X” type. Here, “X” represents a four-digit number in which each successive term is greater than its predecessor. The number of such “1X” type terms is equal to:
$\begin{aligned} & ={}^{8}{{C}_{4}} \\\ & =\dfrac{8!}{4!\times 4!} \\\ & =\dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} \\\ & =70 \\\ \end{aligned}$
Thus, there are 70 terms of the type “1X”.

Now, we will calculate the number of term of the type “2X”. This is equal to:
$\begin{aligned} & ={}^{7}{{C}_{4}} \\\ & =\dfrac{7!}{4!\times 3!} \\\ & =\dfrac{7\times 6\times 5}{3\times 2\times 1} \\\ & =35 \\\ \end{aligned}$
Now, the total number of terms is equal to: 105 $\left( 70+35 \right)$ which exceed 97. Therefore, the first term of our number is 2.
Now, the number of term of the type “23Y”, where “Y” represents a three-digit number is equal to:
$\begin{aligned} & ={}^{6}{{C}_{3}} \\\ & =\dfrac{6!}{3!\times 3!} \\\ & =\dfrac{6\times 5\times 4}{3\times 2\times 1} \\\ & =20 \\\ \end{aligned}$
Now, the number of terms of the type “24Y” is equal to:
$\begin{aligned} & ={}^{5}{{C}_{3}} \\\ & =\dfrac{5!}{3!\times 2!} \\\ & =\dfrac{5\times 4}{2\times 1} \\\ & =10 \\\ \end{aligned}$

Now, the total number of terms is equal to: 100 $\left( 70+20+10 \right)$ which exceed 97. Therefore, the second term of our number is 24. Now, we can just proceed by writing the next numbers in series. These numbers are:
$\begin{aligned} & \Rightarrow {{91}^{th}}\text{ number}=24567 \\\ & \Rightarrow {{92}^{nd}}\text{ number}=24568 \\\ & \Rightarrow {{93}^{rd}}\text{ number}=24569 \\\ & \Rightarrow {{94}^{th}}\text{ number}=24578 \\\ & \Rightarrow {{95}^{th}}\text{ number}=24579 \\\ & \Rightarrow {{96}^{th}}\text{ number}=24589 \\\ & \Rightarrow {{97}^{th}}\text{ number}=24678 \\\ \end{aligned}$
Hence, we can clearly see that the ${{97}^{th}}$ number in the series is equal to 24678 and it does not contain the digits 0, 1, 3, 5 and 9.

Note: Whenever calculating a problem where counting of numbers is required, one should always form a method or technique that can rapidly count these numbers, instead of counting all the numbers from the beginning. The latter may seem easier but it is an ineffective method when finding numbers having greater position in series.