Question

The digits of a nine-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 written in random order, the probability that the number is divisible by 11 is

Answer 1

Hint: First assume any 9 digit number as A, B, C, D, E, F, G, H, I then apply divisibility of 11 rule which says that by adding the alternate digits of number and subtracting each other to find it is divisible by 11 or not. Then use the fact that the sum of digits is 45. Form an inequality using the four digits and then apply the concept of probability to get results.

Complete step by step answer:
In the question we are given digits 1, 2, 3, 4, 5, 6, 7, 8, 9 from which nine digits are formed randomly using all digits.
So, let the number formed be represented as A, B, C, D, E, F, G, H, I is divisible by 11. So we can apply the divisibility rule on it.
Divisibility of 11 can be checked by adding the alternate digits of numbers separately and then subtract it with each other. If the subtracted result is either ‘’$0$” or divided by ‘11’. We can say that the number is divided by 11.
So for A B C D E F G H I, we will check divisibility of S by 11 where S is equal to $\left( A+C+E+G+I \right)-\left( B+D+F+H \right)$
As we know that ABCDEFGHI is formed by digits 1, 2, 3, 4, 5, 6, 7, 8, 9 we can say that sum of the digits is $\left( 1+2+3+4+5+6+7+8+9 \right)$ is equal to 45.
In the ‘S’ we can write $\left( A+C+E+G+I \right)$ as $45-\left( B+D+F+H \right)$ as sum of its digits is 45.
$\Rightarrow $ Is equal to $\left( 45-\left( B+D+F+H \right) \right)-\left( B+D+F+H \right)$
Hence $S=45-2\left( B+D+F+H \right)$
Now let’s take $B+D+F+H$. Its sum will be least when we take values as $1,2,3,4$ so $B+D+F+H$ is $10$. To be the maximum value we can take values as $6,7,8,9$ so $B+D+F+H$ can maximum attain value as 30.
Hence we can write inequality as, $10\le B+D+F+H\le 30$
$\Rightarrow $ We can also rewrite as $-20\ge -2\left( B+D+F+H \right)\ge -60$
Now by adding 45 to all the sides we get, $25\ge 45-2\left( B+D+F+H \right)\ge -15$
As we know that $S=45-2\left( B+D+F+H \right)$ we can write it as $25\ge S\ge -15$
$\Rightarrow $ By reversing we can write it as, $-15\le S\le 25$
$\Rightarrow $ Would be between $-11$ and $11$
For values to be between $-11$ and $11$
$B+D+F+H$ Will be between $17$ and $27$.
Now we have to select a number of ways to choose $4$ from $9$ numbers in $^{9}{{C}_{4}}$ ways.
To find probability we will use formula which is $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where $P\left( E \right)$ is probability of an event, $n\left( E \right)$ number of events which are favorable and $n\left( S \right)$ total number of outcomes
$n\left( E \right)$ Will be $11$ as only $11$ numbers lie between $17$ and $27$.
$n\left( S \right)$ Will be $^{9}{{C}_{4}}$ which is evaluated as $\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}$ which is $126$.
$\Rightarrow P\left( E \right)=\dfrac{11}{126}$.
Hence the probability is $\dfrac{11}{126}$.

Note: Students should know about the concept of permutation and combination also how to find probability and the ways about transforming the inequality.
Sometimes during sign conversion student forget to change the inequality sign, for example,
$1<-x$
Now when we shift the sign, the inequality changes, i.e.,
$-1>x$
If this is not taken care of the answer will be wrong.