Number Selection

[eboolean]

6 random numbers are selected, including 0 and 9. (000000, ..., 999999) Find the probability that the sum of the first three digits of the selected number equals the sum of the last three digits.

 

[Deleted member 4993]

6 random numbers are selected, including 0 and 9. (000000, ..., 999999) Find the probability that the sum of the first three digits of the selected number equals the sum of the last three digits.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this assignment.

The first step would be to calculate the total number of 6 digit numbers (all of the numbers - without any restriction).​

 

[Steven G]

Interesting problem.

Please call the numbers digits--6 random digits are selected to make a 6 digit number.

How many different sums for the 1st three digits? Does each sum have the same probability?

 

[pka]

6 random numbers are selected, including 0 and 9. (000000, ..., 999999) Find the probability that the sum of the first three digits of the selected number equals the sum of the last three digits.

I wonder who wrote this question. It is a counting nightmare.
Consider the six digit number \(\boxed{519}\boxed{078}\) the sum of the first three digits is the sum of the last three.
Any rearrangement of the first three digits, there six, has the same sum. Thus there are thirty six possible six digits with the sum property.
Switch the boxes \(\boxed{078}\boxed{519}\) to get a total of seventy-two. There are \(\dbinom{10}{3}=120\) triples of three distinct digits
That look like \(\boxed{285}\boxed{285}\) and each of those can boxes be rearranged in six ways to total thirty-six.
At this point we have only begun to count.

 

[Steven G]

I thought about this and could only come up with looking at every different possible sum. I am glad that I did not miss anything obvious.

 

[Dr.Peterson]

I thought about this and could only come up with looking at every different possible sum. I am glad that I did not miss anything obvious.

In principle, that's a perfectly reasonable way to solve this. And maybe we can make it easier.

There are 28 possible sums of 3 digits; but since there are as many ways to make 27-n as to make n, we just have to find the number of ways to make the first 14 possible sums from 3 digits. Square each of these to get the number of ways for the first and last 3 digits to both equal this sum; then add up the 14 results and double this to account for the other half of the sums.

So the only hard part is to find the ways to make those 14 sums. You might start by imagining a 9x9x9 cube of possible sums, and then thinking about triangular numbers.

 

[pka]

I thought about this and could only come up with looking at every different possible sum. I am glad that I did not miss anything obvious.

There is nothing obvious about the solution of this question.
I think that the question was written by someone who thought, mistakenly, it was doable with ease.
Please look at this generating polynomial. That tells us that any triple of digits can have twenty-eight different sums, from \(0\) to \(27\).
In the that expansion the term \(73x^{15}\) tells us that there seventy-three ways to get a sum of fifteen.
Two of those triples are \(\boxed{249}~\&~\boxed{366}\). Therefore there are \(73^2\) possible sixtuples of digits having the sum of the first three equal to the sum of the second three. Let's look at the term in the expansion \(3x^{26}\), telling us that there are three ways to get a sum of twenty-six: \(998,~989,~\&~899\).
\(\boxed{998}\boxed{998}\\\boxed{998}\boxed{989}\\\boxed{998}\boxed{899}\\\boxed{989}\boxed{998}\\\boxed{989}\boxed{989}\\\boxed{989}\boxed{899}\\\boxed{899}\boxed{998}\\\boxed{899}\boxed{989}\\\boxed{899}\boxed{899}\)
From which we can see that there are nine six-tuples where the first three & the last three both have a sum of twenty-six.