Question: The sum of the digits of a seven-digit number is 59. Find the probability that this number is divisi...

Question

The sum of the digits of a seven-digit number is 59. Find the probability that this number is divisible by 11.
A) $\dfrac{5}{{21}}$
B) $\dfrac{4}{{21}}$
C) $\dfrac{3}{{22}}$

Answer 1

Solution

Here, we will use the concept of probability and permutation to solve the question. We will first find all the possible seven digit numbers with sum 59. Then we will find the permutation of each number and add them to find a favourable outcome. We will then use the formula of probability to find the required probability.
Formula Used: We will use the following formulas to solve our question:

${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , where $n$ is the total number of things and $r$ is the number of things taken at one time.
${\rm{Probability = }}\dfrac{{{\text{Favorable Outcomes}}}}{{{\text{Total Number of Outcomes}}}}$

Complete step by step solution:
Let us find the various combinations of a seven-digit number that has the sum 59.
The various possible combinations are as follows –
9999995, 9999986, 9999977, 9999887, 9998888
Let us now find the total number of possible permutations for the number 9999995.
We can see that the number 9999995 has a total of seven digits. Out of the seven digits, only 1 digit is unique. And this digit will decide the permutation pattern.
Hence, $n = 7$ and $r = 1$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^7{P_1} = \dfrac{{7!}}{{\left( {7 - 1} \right)!}}\\\ \Rightarrow {}^7{P_1} = \dfrac{{7!}}{{6!}}\end{array}$
On solving the factorial, we get,
$\Rightarrow {}^7{P_1} = 7 \times 1 = 7$
Hence, there are 7 permutations for this.
Let us now find the total number of possible combinations for the number 9999986.
We can see that the number 9999986 has a total of seven digits. Out of the seven digits, 2 digits are unique. And these digits will decide the permutation pattern.
Hence, $n = 7$ and $r = 2$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^7{P_2} = \dfrac{{7!}}{{\left( {7 - 2} \right)!}}\\\ \Rightarrow {}^7{P_2} = \dfrac{{7!}}{{5!}}\end{array}$
On solving the factorial, we get,
$\Rightarrow {}^7{P_2} = 7 \times 6 = 42$
Hence, there are 42 permutations for this.
Let us now find the total number of possible combinations for the number 9999977.
We can see that the number 9999977 has a total of seven digits. Out of the seven digits, only 1 digit is unique but repeating twice. And these digits will decide the permutation pattern.
Hence, $n = 7$ and $r = 2$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^7{P_2} = \dfrac{{7!}}{{\left( {7 - 2} \right)!}}\\\ \Rightarrow {}^7{P_2} = \dfrac{{7!}}{{5!}}\end{array}$
Since the digit 7 is repeating twice, the total number of permutations will be found by dividing with the $2!$ . On doing so, we get,
${}^7{P_2} = \dfrac{{7!}}{{2!5!}}$
On solving this, we get,
$\begin{array}{l}{}^7{P_2} = \dfrac{{7 \times 6}}{2}\\\ \Rightarrow {}^7{P_2} = 7 \times 3 = 21\end{array}$
Hence, there are 21 permutations for this.
Let us now find the total number of possible combinations for the number 9999887.
We can see that the number 9999887 has a total of seven digits. Out of the seven digits, 2 digits are unique but 1 digit is repeating twice. And these digits will decide the permutation pattern.
Hence, $n = 7$ and $r = 3$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^7{P_3} = \dfrac{{7!}}{{\left( {7 - 3} \right)!}}\\\ \Rightarrow {}^7{P_3} = \dfrac{{7!}}{{4!}}\end{array}$
Since the digit 8 is repeating twice, the total number of permutations will be found by dividing with the $2!$ . On doing so, we get,
${}^7{P_3} = \dfrac{{7!}}{{2!4!}}$
On solving this, we get,
$\begin{array}{l}{}^7{P_3} = \dfrac{{7 \times 6 \times 5}}{2}\\\ \Rightarrow {}^7{P_3} = 7 \times 3 \times 5\\\ \Rightarrow {}^7{P_3} = 105\end{array}$
Hence, there are 105 permutations for this.
Let us now find the total number of possible combinations for the number 9998888.
We can see that the number 9998888 has a total of seven digits. Out of the seven digits, only 1 digit is unique but it is repeating four times. And these digits will decide the permutation pattern.
Hence, $n = 7$ and $r = 4$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^7{P_4} = \dfrac{{7!}}{{\left( {7 - 4} \right)!}}\\\ \Rightarrow {}^7{P_4} = \dfrac{{7!}}{{3!}}\end{array}$
Since the digit 8 is repeated four times, the total number of permutations will be found by dividing with the $4!$ . On doing so, we get,
${}^7{P_4} = \dfrac{{7!}}{{4!3!}}$
On solving this, we get,
$\begin{array}{l}{}^7{P_4} = \dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}\\\ \Rightarrow {}^7{P_4} = 7 \times 5\\\ \Rightarrow {}^7{P_4} = 35\end{array}$
Hence, there are 35 permutations for this.
Thus, the total number of permutations will be,
Total number of permutations $= 7 + 42 + 21 + 105 + 35$
Total number of permutations $= 210$
Hence, the total number of permutations is 210.
We will now find those numbers from the above permutations, that are divisible by 11.
We will use the Divisibility test of 11 here.
Now the odd number of digits should add up to 35 here.
So, they will be 9998.
Let us now find the total number of possible permutations for the number 9998.
We can see that the number 9998 has a total of four digits. Out of the four digits, only 1 digit is unique. And this digit will decide the permutation pattern.
Hence, $n = 4$ and $r = 1$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^4{P_1} = \dfrac{{4!}}{{\left( {4 - 1} \right)!}}\\\ \Rightarrow {}^4{P_1} = \dfrac{{4!}}{{3!}}\end{array}$
On solving the factorial, we get,
${}^4{P_1} = 4 \times 1 = 4$
Hence, there are 4 permutations for this.
Now the even number of digits should add up to 24 here.
So, they will be 996, 987, and 888.
Let us now find the total number of possible permutations for the number 996.
We can see that the number 996 has a total of three digits. Out of the three digits, only 1 digit is unique. And this digit will decide the permutation pattern.
Hence, $n = 3$ and $r = 1$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^3{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}}\\\ \Rightarrow {}^3{P_1} = \dfrac{{3!}}{{2!}}\end{array}$
On solving the factorial, we get,
$\Rightarrow {}^3{P_1} = 3 \times 1 = 3$
Hence, there are 3 permutations for this.
Let us now find the total number of possible permutations for the number 987.
We can see that the number 987 has a total of three digits. Out of the three digits, two digits are unique. And these digits will decide the permutation pattern.
Hence, $n = 3$ and $r = 2$ . On substituting these values in the formula for the permutation, we get,
$\begin{array}{l}{}^3{P_2} = \dfrac{{3!}}{{\left( {3 - 2} \right)!}}\\\ \Rightarrow {}^3{P_2} = \dfrac{{3!}}{{1!}}\end{array}$
On solving the factorial, we get,
$\Rightarrow {}^3{P_2} = 3 \times 2 = 6$
Hence, there are 6 permutations for this.
Let us now find the total number of possible permutations for the number 888.
We can see that the number 888 has a total of three digits. All three digits are the same.
Hence, there is only 1 permutation for this.
So, the total number of permutations for the even numbers will be,
Total number of permutations for even numbers $= 3 + 6 + 1$
Total number of permutations for even numbers $= 10$
So, the total number of ways in which a permutation is divisible by 11, will be,
We will use the multiplication rule of permutations here.
Total number of permutations for a number to be divisible by 11 $=$ Total number of permutations of odd numbers $\times$ Total number of permutations of even numbers
On substituting the values, we get,
Total number of permutations for a number to be divisible by 11 $= 4 \times 10$
Total number of permutations for a number to be divisible by 11 $= 40$
Thus, our total number of outcomes will be the total number of permutations for a seven-digit number having sum 59, which is 210.
Thus, our favorable number of outcomes will be the total number of permutations for a seven-digit number having sum 59 and divisible by 11, which is 40.
Let us calculate our probability now.
On substituting the above values in the formula for probability we get,
$\begin{array}{l}{\rm{Probability = }}\dfrac{{{\text{Favorable Outcomes}}}}{{{\text{Total Number of Outcomes}}}}\\\ \Rightarrow {\rm{Probability = }}\dfrac{{40}}{{210}}\\\ \Rightarrow {\rm{Probability = }}\dfrac{4}{{21}}\end{array}$
Thus, our required probability is ${\rm{ }}\dfrac{4}{{21}}$ .

So, the correct answer is option (B).

Note:
We have used the divisibility test of 11 here. The divisibility test of 11 states that a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number which is divisible by 11.
Example – Consider the number 29435417.
Sum of its digits at odd places $= 7 + 4 + 3 + 9$
Sum of its digits at odd places $= 23$
Sum of its digits at even places $= 1 + 5 + 4 + 2$
Sum of its digits at even places $= 12$
Now, according to the divisibility test,
$23 - 12 = 11$
And 11 is divisible by 11, so it is true.