# Math Help - prob 1

1. ## prob 1

seven digits from the digits 1 2 3 4 5 6 7 8 9 are written in a random order.the probability that this seven digit no is divisible by 11 is

how to do this
i have got total cases as 9*8*7*6*5*4*3

2. ## Re: prob 1

If a 7 digits no. "a b c d e f g" is divisible by 11, then

a+c+e+g = b+d+f

But I don't know the reason and the method to acquire that combinations.

3. ## Re: prob 1

Hello, prasum!

Seven digits from the digits {1, 2, 3, 4, 5, 6, 7, 8, 9} are written in a random order.
Find the probability that this seven-digit number is divisible by 11.

i have got total cases as: 9*8*7*6*5*4*3 = 181,440 . Right!

BookEnquiry has the right idea . . .

Let $ABCDEFG$ be the 7-digit number.

If $ABCDEFG$ is a multiple of 11,
. . then $(A+C+E+G)-(B+D+F)$ is a multiple by 11.

I found no formula for counting the multiples-of-11.
I was forced to use Brute Force listing.

I found 54 sets of digits which produce multiples-of-11.

$\begin{array}{cc}\text{A, C, E, G} & \text{B, D, F} \\ \hline 9,8,7,6 & 5,2,1 \\ 9,8,7,6 & 4,3,1 \\ 9,8,7,5 & 4,2,1 \\ 9,8,7,4 & 3,2,1 \\ 9,8,7,2 & 6,5,4 \\ 9,8,7,1 & 6,5,3 \\ 9,8,6,5 & 3,2,1 \\ 9,8,6,2 & 7,4,3 \\ 9,8,6,1 & 7,4,2 \\ 9,7,6,4 & 8,5,2 \\ 9,7,6,3 & 8,5,1 \\ 9,7,6,3 & 8,4,2 \\ 9,6,5,4 & 8,3,2 \\ 9,6,5,3 & 7,4,1 \\ 9,6,5,2 & 7,3,1 \\ 9,5,4,3 & 7,2,1 \\ 9,4,3,1 & 8,7,2 \\ 9,3,2,1 & 6,5,4 \end{array}$ . . $\begin{array}{cc} \text{A, C, E, G} & \text{B, D, F} \\ \hline 8,7,6,4 & 9,3,2 \\ 8,7,6,2 & 5,4,3 \\ 8,7,6,1 & 5,4,2 \\ 8,7,5,4 & 9,3,1 \\ 8,7,5,3 & 9,2,1 \\ 8,7,5,3 & 6,4,2 \\ 8,7,5,2 & 6,4,1 \\ 8,6,5,4 & 9,2,1 \\ 8,6,5,4 & 7,3,2 \\ 8,6,5,1 & 4,3,2 \\ 8,5,4,2 & 9,7,3 \\ 8,5,4,1 & 9,7,2 \\ 8,5,4,1 & 9,6,3 \\ 8,4,3,2 & 9,7,1 \\ 8,4,3,1 & 9,5,2 \\ 7,6,5,4 & 8,2,1 \\ 7,6,5,3 & 9,8,4 \\ 7,6,5,2 & 9,8,3 \end{array}$ . . $\begin{array}{cc}\text{A, C, E, G} & \text{B, D, F} \\ \hline 7,6,5,1 & 9,8,2 \\ 7,5,4,3 & 9,8,2 \\ 7,5,4,2 & 9,8,1 \\ 7,5,4,2 & 9,6,3 \\ 7,5,4,1 & 9,6,2 \\ 7,4,3,2 & 9,6,1 \\ 7,4,3,1 & 8,5,2 \\ 6,5,4,3 & 9,8,1 \\ 6,5,4,3 & 9,7,2 \\ 6,5,4,2 & 9,7,1 \\ 6,5,3,2 & 8,7,1 \\ 6,5,3,1 & 9,4,2 \\ 6,4,3,2 & 9,5,1 \\ 6,4,3,1 & 7,5,2 \\ 5,4,3,2 & 7,6,1 \\ 5,3,2,1 & 9,7,6 \\ 4,3,2,1 & 9,7,5 \\ 4,3,2,1 & 8,7,6 \end{array}$

In each of the 54 sets, $\{A,C,E,G\}$ can be arranged in $4!$ ways
. . and $\{B,D,F\}$ can be arranged in $3!$ ways.

Hence, there are: . $54 \times 4!\times 31 \:=\:7,\!776$ numbers divisible by 11.

Therefore: . $P(\text{div. by 11}) \;=\;\frac{7,\!776}{181,\!440} \;=\;\frac{3}{70}$