Question

Seven digits from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in random order. The probability that this seven-digit number is divisible by 9 is\[\dfrac{k}{9}\] find k?

Answer 1

Hint:
 In this problem, seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are randomly arranged. To find the value of k, where the probability that this seven-digit number is divisible by 9 is \[\dfrac{k}{9}\]. We know that the integer is divisible by 9 if the sum of its digits is divisible by 9. Find the sum of its digits and check for divisibility. Then it must be the case that the sum of two integers we don’t pick is divisible by 9 to form the seven-digit number. There will be four two-digit pairs whose sum is divisible by 9. Take the ratio to get the probability that the seven-digit number formed is divisible by 9.

Complete step by step answer:
We know that,
If the number is divisible by 9, the sum of the digits of that number will also be divisible by 9.
Some of the first nine natural numbers= 45
Here, 45 is divisible by 9.
We know that, if we select seven numbers so that the combination of these numbers is divisible by 9, then the sum of the remaining two numbers is divisible by 9.
So, now we can choose the two integers, whose sum is divisible by 9.
Number of ways of choosing two integers from 9 integers in combination method, we get
Total case = \[{}^{9}{{C}_{2}}\]= 36
There are four of two-digit pairs whose sum is divisible by 9,
The favorable cases are: {(1, 8), (2, 7), (3, 6), (4, 5)}.
Here, the favorable cases = 4 and the Total case = 36.
Now we are going to take the ratio to get the probability that the seven-digit number so formed is divisible by 9.
By dividing the Favorable case by the total case, we will get the probability.
\[\Rightarrow \dfrac{4}{36}=\dfrac{1}{9}\]
Therefore, the value of k is 1.

Note:
  Students will make mistakes while choosing the number of favorable cases, that is the two-digit pair whose sum is divisible by 9. Another method:
The total number of seven-digit number is
\[{}^{9}{{C}_{2}}={}^{9}{{C}_{7}}=36\]
For divisibility with 9, leave one of (1, 8), (2, 7), (3, 6), (4, 5)’
Here we got 4 pairs, that is four favorable cases, which must be divided by the total case to find the probability
Therefore, the required probability = \[\dfrac{4}{36}=\dfrac{1}{9}\]
 \[\therefore k=1\]