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We have a sequence of $10$ digits, how many ways are there that the sum of all the digits results into an even number?

Examples: $2464244482$, the sum of all the digits is an even number.

My Attempt

I listed out a few trails and saw that I need a even number of odd digits, and a even number of even digits.

$$\begin{aligned} \text{Odd} && \text{Even} \\ 0 && 10 \\ 2 && 8 \\ 4 && 6 \\ 6 && 4 \\ 8 && 2 \\ 10 && 0 \end{aligned}$$ Now the problem is I don't know how to put this into a combinations and permutations setting: My initial thought was: $$\binom{5}{1}^{10} + \binom{5}{1}^{8}\binom{5}{1}^{2} + \binom{5}{1}^{6}\binom{5}{1}^{4} + \binom{5}{1}^{4}\binom{5}{1}^{6} + \binom{5}{1}^{2}\binom{5}{1}^{8} + \binom{5}{1}^{10}$$ The English to what is above, "Choose an even digit from the group of even digits and we will do this $10$ times, or we choose an even number $8$ times then choose an odd number $2$ times." We keep doing this until we have gone through all the cases.

Is this a correct way of doing this, and is there a more simple way of solving this?

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    $\begingroup$ Pick the first nine digits in a random manner, let us fix such a configuration. Consider the event $A$ of all possibilities to pick now the tenth digit. In the cases $0,2,4,6,8$ we have one parity, in the other cases, $1,3,5,7,9$ the other parity. So constrained by $A$, the (conditional) probability to get an even number is $1/2$. So the wanted probability is $1/2$... $\endgroup$
    – dan_fulea
    Commented Sep 7, 2022 at 16:55
  • $\begingroup$ @dan_fulea Except that leading zeros are usually not allowed. $\endgroup$
    – Peter
    Commented Sep 7, 2022 at 16:56
  • $\begingroup$ @Peter pick them so that they are allowed. (So do not pick the first one randomly among $0,1,2,3,4,5,6,7,8,9$, the second one... but pick them as a whole from the set of all possible first nine digits.) $\endgroup$
    – dan_fulea
    Commented Sep 7, 2022 at 16:59
  • $\begingroup$ @Peter, leading $0$s are allowed. Should of put in OG question. $\endgroup$
    – James
    Commented Sep 7, 2022 at 17:05
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    $\begingroup$ Fixing your attempt, $5^{10}+\binom{10}{2}5^{10}+\binom{10}{4}5^{10}+\binom{10}{6}5^{10}+\binom{10}{8}5^{10}+5^{10} = 5^{10}\cdot (\binom{10}{0}+\binom{10}{2}+\dots+\binom{10}{10}) = 5^{10}\cdot 2^9$ where the last simplification relies on this. $\endgroup$
    – JMoravitz
    Commented Sep 7, 2022 at 18:03

1 Answer 1

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(The problem was solved but in comments)


You'll note that everything ultimately matters on the last digit that you choose.

For any $10$-digit number that you take, the first $9$ numbers will return either an Even or an Odd sum and correspondingly you'll then be left with $5$ choices to choose your final, that is, the $10^{th}$ digit.

Suppose the sum of the $1^{st}$ $9$ digits is Odd then you must choose a number from $\{1,3,5,7,9\}$ to get an even sum.
Suppose the sum of the $1^{st}$ $9$ digits is Even then you must choose a number from $\{0,2,4,6,8\}$ to get an even sum.

Now we simply have to count the number of ways of choosing the numbers.
For each of the $1^{st}$ $9$ digits, we have $10$ choices - $\{0,1,2,3,4,5,6,7,8,9\}$
But the last last digit can only take $5$ values for whether we have an Odd or an Even sum of the first $9$ digits to get a resultant even sum.
Thus, the total number of ways that the sum of all the digits results into an even number is $$\boxed{5\cdot10^9}$$

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