Probability divisibility test

There are 9! different combinations of the 9 digits, or 362880 possibilities. This can easily be churned through by a computer.

Python 3.1 code:

from itertools import permutations
from fractions import Fraction

def to_num(iterable):
    val = next(iterable, None)
    return 0 if val == None else val + 10 * to_num(iterable)

p, q = 0, 0

for digits in permutations(range(1, 10)):
    if to_num(reversed(digits)) % 11 == 0: p += 1
    q += 1

print(Fraction(p, q))

This prints "11/126" which sums to 137.

1+2+3+4+5+6+7+8+9 = 45. To get a nine non-zero distinct digits number which is also divisible by eleven, we need the difference of odd and even placing of digits to be a multiple of 11. 45 is odd, so the difference cannot be zero nor it any more than 11, or else we have to find at least four non-zero distinct digits to add up to as low as (45-3x11)/2 = 6, which is impossible (1+2+3+4 = 10). If the even-odd difference be 11, then the lesser side should sum at (45-11)/2 = 17.
17 = 1+2+3+4+7.
= 1+2+3+5+6.
= 1+2+5+9.
= 1+2+6+8.
= 1+3+4+9.
= 1+3+5+8.
= 1+3+6+7.
= 1+4+5+7.
= 2+3+4+8.
= 2+3+5+7.
= 2+4+5+6.
So as listed above, there are eleven ways to do groupings, with the former two forming the odd placing and latter nine the even ones.
P(divisible by 11).
= (11 groupings) (5! of odd placing) (4! of even placing) / (9! total freedom of digits placing).
= 11/126

Let the sum of the digits at even places be $x$ and sum of digits at odd places be $y$ .

Then $x+y=45$

For the number to be divisible by 11 , $x-y = 11z$ where z is an integer. $z$ can be 1 or -1.

Possible values are $(x,y)=(17,28),(28,17)$

Number of such $x$ and $y$ can be calculated and comes out to be 11.

Therefore Probability = $\dfrac{4!\times5!\times11}{9!}$ = $\dfrac{11}{126}$

$p+q= \boxed{137}$

The result is (11*(4!)(5!))/(9!)

The sample space can be easily found out: 9!

Now for the favourable cases, the divisibility test of 11 is: the difference of ( sum of digits at even places and sum of digits at odd places) must be divisible by 11:

And since the 9! < 10^6. We can do it by a program.

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#AnubhavBalodhi, Brilliant, level 5 in Combinatronics, Probability divisibility test
import itertools
from math import factorial
from fractions import Fraction
array=list(range(1,10))
for num in array:
    print(num)
A=list(itertools.permutations((array)))
ans=0
for a in A:
    odd=0
    even=0
    i=0
    for num in a:
        if i&1:
            odd+=num
        else:
            even+=num
        i+=1
    if (even-odd)%11 ==0:
        #print(a)
        ans+=1

print(ans)

a=ans
def gcd(a,b):
    if(b==0):return a;
    return gcd(b,a%b);

b=factorial(9)
c=gcd(a,b)
ANS=Fraction(a,b)
ans=int(a/c)+int(b/c)
print(a,b,c,int(a/c),int(b/c),ans,ANS)

It gives ans=137.