How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even? The task is the following:
$M= \left \{ 1,2, ... 99,100 \right \}$
How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even?
I tried to solve it this way:
There are only two possibilities that $a+b+c$ is even:
First I look at this as a unordered set:
Order isn't important  and you can put something back gives me:
one element is even and the other two are not: 
$$\binom{50+1-1}{1}\binom{50 +2 - 1}{2}$$
or all three elements are even which gives me:
$$\binom{50+3-1}{3}$$
Now I add both and multiply it by $3!$ because there are $6$ possibilities for an unordered set to be ordered and order is important in tuples.
I get 
$$3! \cdot \left[\binom{50 +2}{3} + 50 \binom{50 +1}{2}\right] = 515. 100$$
I think the solution is $500.000$ isn't it? Can't find my mistake...
 A: Total number of tuples - $100^3$. Half of them are even, half of them are odd by symmetry. So, answer is $500,000$. (Remember!! $a,b,c$ can be equal) Proof for symmetry? Well, see this- For every sum which is odd, there exists a sum which is even, obtained by adding $1$ to one of the numbers contributing to the odd sum! eg: If you give me $100+100+99 = 299$, I give you $100+100+100=300$ 
similarly, prove the inverse- For every sum which is even, there exists a sum which is odd, obtained by subtracting $1$ from one of the numbers contributing to the even sum!
A: You could just list the permissible odd-even configurations as:
$OEE,\;\; EOE,\;\; EEO,\;\;$ and $\;\;EEE\;\;$  each with $50^3$ possibilities,
thus ans $=4\times50^3$  
A: Your assumption that there are $3!$ of arranging each ordered triple $(a, b, c)$ is only valid if $a$, $b$, and $c$ are distinct.  However, $(a, b, c) \in M^3$, so $a$, $b$, and $c$ need not be distinct.  
Let's consider cases.
Three even numbers are selected:


*

*There are $50$ ordered triples of the form $(a, a, a)$.

*Ordered triples $(a, b, c)$ in which exactly two of the numbers are equal.  We have $50$ choices for the repeated number, $\binom{3}{2}$ choices for their places in the triples, and $49$ choices for the other even number, so there are $$50 \cdot 49 \cdot \binom{3}{2}$$ such triples.

*Ordered triples $(a, b, c)$ in which each even number is distinct.  There are $50$ choices for $a$, $49$ choices for $b$, and $48$ choices for $c$.  Hence, there are $50 \cdot 49 \cdot 48$ of these.


One even number and two odd numbers are selected:


*Ordered triples with one even number and a repeated odd number.  There are $50$ choices for the even number, $3$ choices for its location, $50$ choices for the repeated odd number, and one way of placing the repeated odd number in the open locations.  Hence, there are $50^2 \cdot 3$ of these.

*Ordered triples with one even number and two distinct odd numbers.  There are $50$ choices for the even number, $\binom{50}{2}$ choices for the two odd numbers, and $3!$ permutations of the three distinct numbers.  Hence, there are $$50 \cdot \binom{50}{2} \cdot 3!$$ of these.
Since the cases are mutually exclusive, the total number of ways in which the numbers can be selected is 
$$50 + 50 \cdot 49 \cdot \binom{3}{2} + 50 \cdot 49 \cdot 48 + 50^2 \cdot 3 + 50 \cdot \binom{50}{2} \cdot 3! = 500,000$$
A: The problem is tuples with repeated numbers.  If you've counted $(2,4,4)$ as a tuple in the first stage, then there are only $3$ (not $3!$) permutations of the unordered numbers.
If it helps, there are $125.000$ combinations with all $a,b,c$ even, and $375.000$ with two odd and one even.
