Question

A seven digit number is formed by using 0,1,2,3,4,8,9 without repetition. Then what is the probability that it will be divisible by 4?
(a) $\dfrac{53\times 4!}{5\times 6!} $
(b) $\dfrac{53}{180} $
(c) $\dfrac{53!}{6\times 6!} $
(d) None of these

Answer 1

Hint: You should note that two or more than two digit numbers are divisible by 4 if it’s last two digits of the number are divisible by 4. So, you should try to find the number of such numbers whose last two digits are divisible by 4 and then divide it by the total number of numbers that can be formed using the given digits to find the probability. You should try to form cases involving 0 and without 0 like when 0 is in the last two digits then five cases are possible 04, 08, 20, 40, 80, and first five digits can be arranged in 5! Ways so the total number of favourable cases are 5 $\times $ 5!, similarly we can make other cases also.

Complete step-by-step answer:
We should know that two or more than two digit numbers are divisible by 4 if it’s last two digits of the number are divisible by 4,
Digits that are given to us are 0,1,2,3,4,8,9 so possible last two digits can be,
04, 08, 12, 20, 24, 28, 32, 40, 48, 80, 84, 92.

Case 1: we will suppose that 0 is in last two digits of the number, then:
Different cases that will be possible are 04, 08, 20, 40, 80, hence we get
Total number of different cases will be 5 when 0 included in last two digits, and
We know that, first five digits can be arranged in 5! Ways, hence, we get
Total number of ways to form a number according to case 1 will be = 5 $\times $ 5!

Case 2: we will suppose when 0 in neither at last two digits and nor at first place then,
There are four places where 0 can be placed i.e. are from ${{2}^{nd}} $ to ${{5}^{th}} $ so, we get
Total number of ways zero can be placed will be = 4
And we can conclude that,
last two digits can be placed in 7 ways = 12, 24, 28, 32, 48, 84, 92
Out of the first five digits four digits can be arranged in 4! Ways (because 0 is excluded), and we get
Total number of cases according to case 2 will be = 4 $\times $ 4! $\times $ 7

And also we will find out total number of 7 digit numbers that can be formed using given digits will be,
= 7! – 6! = 6 $\times $ 6!
As 6! Is subtracted from 7! As we considered a case when 0 can’t be placed at first placed and in that case number of cases will be 6! otherwise 7! cases are there for 7 numbers.
Now, we know
Probability = $\dfrac{favourable\,cases}{total\,number\,of\,cases}=\dfrac{case1+case2}{total\,number\,of\,cases} $
= $\dfrac{(5\times 5!)+(4\times 7\times 4!)}{6\times 6!} $
After further solving the above expression, we get
Probability = $\dfrac{53}{180} $
So, the correct answer is “Option ”.B

Note: You need to pay special attention if in these types of questions 0 is one of the given digits, Whenever 0 is given try to break the given problem to sub-parts involving 0 at first place and likewise when it is not at first place. Most of the students also make mistakes in finding the total number of cases they just write it as 7! And not subtract the case when 0 is at first place, hence always be extra cautious while making cases which involve 0 as one of its digits.