Question

All the five digit numbers N = abcde having property a < b < c < d < e are arranged in the increasing order of their magnitude. Then the ${97}^{th}$number in the list contains the odd digit is

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

The correct option is C

7

To make five digits numbers out of the numbers from 1, 2, 3, 4, ......7, 8, 9:
Case 1: The five digit number starting with 1:

The total number of ways for selecting remaining 4 places out of total digits 2, 3, 4, 5, 6, 7, 8, 9 is ${}^8{C}_4=\frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=70$ ways.
Case 2: The five digit number starting with 2:

The total number of ways for selecting remaining 4 places out of the digits 3, 4, 5, 6, 7, 8, 9 is ${}^7{C}_4=\frac{7\times 6\times 5\times 4}{4\times 3\times 2\times 1}=35$
In both the cases the total five digits numbers is $70+35=105$
Since, we required 97th number, so case 2, will not help to get the required five digit number.
Case 3: The five digit number starting with 2 and 3:

The number of ways to select remaining three position out of the digit 4, 5, 6, 7, 8, 9 is ${}^6{C}_3=\frac{6\times 5\times 4}{3\times 2\times 1}=20$
Thus, the total number of five digits number in case 1 and case 3 is $70+20=90$
Now, still 6 more five digits numbers, we need to find to get the 97th number in the list.
Case 4: The five digit number start with 2 , 4 and 5 is

The number of ways to select 2 positions out of 6, 7, 8 and 9 is${}^4{C}_2=\frac{4\times 3}{2\times 1}=6$
Thus, the total number of 5 digits number in case 1 , case 3 and case 4 is $70+20+6=96$
Thus, the next immediate least number is 24678, which is 97th number in list, so it contains an odd digit of 7.