Question

An eight-digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. What will be the number of ways in which this can be done?
A. $40\left( {7!} \right)$
B. $36\left( {7!} \right)$
C. $18\left( {7!} \right)$
D. $72\left( {7!} \right)$

Answer 1

Hint: In this question, use the concept of divisibility check of number 9. A number is divisible by 9 if the sum of its digits is also divisible by 9. 10 digits are given in question, eliminate a pair of digits from 10 digits to form an 8-digit number such that it is also divisible by 9. Simply calculate the total number of ways in which it is done.

Complete step-by-step answer:
Given digits $ = \left\{ {0,1,2,3,4,5,6,7,8,9} \right\}$ i.e. 10 digits
Sum of digits $ = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ which is divisible by 9 because when 45 is divided by 9
\[
  9\mathop{\left){\vphantom{1{45}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{45}}}}
\limits^{\displaystyle \,\,\, 5} \\
   - {\text{ }}\underline {45} \\
  {\text{ 0}} \\
\]

Since, remainder is equal to 0 so we can say that 45 is divisible by 9.
Eliminating a pair of digits such that 8-digit number is also divisible by 9.
For this the eligible pairs are $ = (0,9),(1,8),(2,7),(3,6),(4,5)$ {as sum of these digits also equal to 9}
After eliminating these the sum will also be divisible by 9.
Eliminating pair $ = (0,9)$
The number of ways in which 8-digit number can be formed (digits without repeating) $ = 8!$
Eliminating pair $ = (1,8)$
The number of ways $ = 7 \times 7!$ because the digit 0 cannot occupy first place
Eliminating pair $ = (2,7)$
The number of ways $ = 7 \times 7!$
Eliminating pair $ = (3,6)$
The number of ways $ = 7 \times 7!$
Eliminating pair $ = (4,5)$
The number of ways $ = 7 \times 7!$
Now, Total number of ways in which it can be done \[ = 4 \times 7 \times 7! + 8! = 28 \times 7! + \left( {8 \times 7!} \right) = 36 \times 7!\]
Hence, the number of ways by which an eight-digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits is $36(7!)$
$\therefore $ Option B. $36(7!)$ is our correct answer.


Note- In such questions, just know the concept of divisibility by 9 and also use the concept of selection and combination to find the total number of ways the required process can be done by simplifying the expression and matching that with the option. We can also approach the problem in such a way that when (1,8) is removed we can simply say that the number of 8-digit numbers $ = 8! - 7!$ (subtracting the number of cases where '0' is at the left most place). Similarly, when we remove (2,7), (3,6) and (4,5) we get $8! - 7!$ in each case. So, Total 8-digit numbers $ = 8! + 4(8! - 7!) = 5 \times 8! - 4 \times 7! = 40 \times 7! - 4 \times 7! = 36 \times 7!$