Question

The number of all ten digit numbers with all the digits distinct and which are divisible by 4 is (if repetition of digits not allowed)
A) $2 \cdot \dfrac{{10!}}{9}$
B) $6 \times 8! + 112 \times 7!$
C) $25 \times 8!$
D) All of these

Answer 1

Here, we will find the number of all ten digit numbers having 0 in the last 2 digits. Then we will find the number of all ten digit numbers having numbers other than 0 in the last 2 digits. Then we will add both the number of ten digit numbers with all the digits distinct and simplify further using the concept of factorial to get the required answer.

Complete step by step solution:
We know that if the last two digits of a number is divisible by 4, then the whole number is divisible by 4.
The total number of two digit numbers divisible by 4 is 24 of which 44 and 88 are not allowed since the repetition of digits is not allowed.
Thus, the total number of two digit numbers divisible by 4 is 22.
The two digit number which contains zero in the ten digit number is 6.
So, the number of two digit numbers with zero in the two digit number is 6.
Thus, the number of last two digits with a non- zero digit $= 22 - 6$
Subtracting the terms, we get
$\Rightarrow$ The number of last two digits with a non- zero digit $= 16$
Now, we will find the number of ways that the two digit number is zero.
If the last two digit number will be a number with the number which contains zero in it, then it will be among the 6 numbers with zeros in it. The remaining digits in a ten digit number will be the 8 distinct digits.
Number of ways that the last two digit number contains zero $= 8! \times 6$
We know that if the last two digit number will be a number with the number which contains non- zero digit in it, then it will be among the 16 numbers with non- zero digit in it.
The remaining digits in a ten digit number will be the 8 distinct digits. These 16 numbers can be arranged in 7 different ways.
Number of ways that the last two digit number which contains non- zero digit $= 16 \times 7 \times 7!$
Multiplying the terms, we get
$\Rightarrow$ Number of ways that the last two digit number which contains non- zero digit $= 112 \times 7!$
Now, we have to find the total number of ways that the ten digit number is divisible by 4, so we will add the total number of ways obtained above. Therefore,
Total Number of ten digit numbers with all the digits as distinct $= 6 \times 8! + 112 \times 7!$
Therefore, the number of all ten digit numbers with all the digits distinct and which are divisible by 4 is $6 \times 8! + 112 \times 7!$.

Thus, option (B) is the correct answer.

Note:
We should be careful that the number of digits should not allow repetition. Repetition of digits is defined as the repeating set of digits. We should remember that it is enough to find the divisibility of 4 by considering only the last two digits of a number. Also, there is a possibility that the last two digits may be zero or a non- zero number.