probability help required please..

Question

seven digit number  from the digits   1,2,3,4,5,6,7,8,9  are written in a random  order .The probability that this seven digit number is divisible by 9 is?

 

1. 1/9
2. 1/3
3. 1/5
4. 2/9

Answer 1

interesting problem; I know it's already been done but here's a solution...

digits 1,2,3,4,5,6,7,8,9

pick 7 digits from the 9 digits

9*8*7*6*5*4*3=181,440 7-digit numbers altogether

how many ways can you arrange the 7-digit numbers ?

7*6*5*4*3*2*1=5040 ways

to get a multiple of 9 you can only eliminate 1 and 8, 2 and 7, 3 and 6, or 4 and 5 because all the digits from 1-9 add up to 45

5040 arrangements without 1 and 8

5040 arrangements without 2 and 7

5040 arrangements without 3 and 6

5040 arrangements without 4 and 5

there are four 5040 arrangements for a total of 4*5040=20,160 arrangements that are divisible by 9

20,160/181,440=2016/18,144=504/4536=126/1134=14/126=7/63=1/9

20,160/181,440=1/9

 

 

 

 

 

Answer 2

In only a few seconds, my computer counted from 1,111,111 to 9,999,999 and found 531,441 numbers which were evenly divisible by 9 and 4,251,528 which were not (only 9^7 numbers to evaluate).  This is a ratio of 1:8 or a probability of 1 out 9.  So, let's determine that mathematically (who needs a computer, anyway? -- just slow and error-prone people like me).
 
"If the sum of the digits of a number is evenly divisible by 9, then the number is evenly divisible by 9."  (a rule; proof not included here)
 
With 7 digits chosen from 1,2,3,4,5,6,7,8,9 and allowing duplicates, the sum of the digits ranges from 7*1 to 7*9 (63 values).  Now, even though permutations of the seven digits determine the frequency of each sum, we don't care -- there are only 63 possible sums of the digits.
 
The order of the digits in the original number doesn't matter, we only care whether they add up to a value evenly divisible by 9.
 
Out of the 63 sums, how many are a multiple of 9?  --  9, 18, 27, 36, 45, 54, 63  -- there are 7,
 
So, 7 out of 63 is 1 out of 9 or  1/9.

Answer 3

What is the divisibility rule for 9?

If you can add up the digits and the sum is divisible by 9, then the original answer is divisible by 9.

 

For example, 7,862,571: 7+8+6+2+5+7+1=36...........36:3+6=9.............9 is divisible by 9, so our original number is divisible by 9.

 

But this mean that 1,752,687 is also divisible by 9, because of the Commutative Property of Addition.

So does order of our numbers matter in this question?

Nope.

 

So if order does NOT matter we use nCr.

nCr=     n!    

         r! (n-r)!

 

We have 9 digits and we need to choose 7 of them:

 

9C7 =      9!    = 9*8*(7!) = 9*8 = 36

          7! (9-7)!   (7!)(2!)        2

 

How many ways can you choose 7 digits from the 9?

 

This will give you the DENOMINATOR of your probability.... the total number of ways you can choose 7 digits.

 

Now to get the NUMERATOR of your probability, we must figure out how many of those combinations are divisible by 9.  It's a small enough list that you can write it systematically and see what you find...

 

1+2+3+4+5+6+7=28 NO

1+2+3+4+5+6+8=29 NO

1+2+3+4+5+6+9=30 NO

1+2+3+4+5+7+8=30 NO

1+2+3+4+5+7+9=31 NO

1+2+3+4+6+7+8=31 NO

etc.... this should take you no more than 2 minutes.  However many of these sums ARE divisible by 9, that is the numerator of your probability.

 

Then reduce your fraction for good measure, or change to a decimal or percent, and you're set!