How many triplets $(x,y,z)$ can we make with $x,y,z\in\{1,\ldots,25\}$, where $x\leq y\leq z$? The numbers $x$, $y$ and $z$ are chosen from the set of $\{1,2,3,\ldots,25\}$ such that $z\geq y\geq x$. In how many different ways can we from such triplets?
 A: Hint: consider the set of triplets $(x,y+1,z+2)$ with $1\leq x<y+1<z+2\leq27$.
A: $3\;\;of\; a\; kind:\;\; \binom{25}1 = 25$
$2-1\;\; of\; a\; kind:\;\; \binom{25}1\binom{24}1 = 600$
$1-1-1\;\; of\; a\;\; kind:\;\; \binom{25}{3} = 2300$
Add up
A: A particular selection is completely determined by the number of times each number is selected.  For instance, if we choose two $7$'s and an $18$, then $x = y = 7$ and $z = 18$.  Let $x_k$, $1 \leq k \leq 25$, denote the number of times the number $k$ is selected.  Since we are selecting a total of three numbers, 
$$x_1 + x_2 + x_3 + \cdots + x_{25} = 3$$
This is an equation in the non-negative integers.  A particular solution corresponds to the placement of $24$ addition signs in a row of $3$ ones.  For instance, 
$$+ + + + + + 1 1 + + + + + + + + + + + 1 + + + + + + +$$
corresponds to the solution $x = y = 7$ and $z = 18$.  Therefore, the number of solutions of the equation is the number of ways $24$ addition signs can be inserted in a row of three ones, which is 
$$\binom{24 + 3}{24} = \binom{27}{24} = \binom{27}{3} = \binom{24 + 3}{3}$$
since we must choose which $24$ of the $27$ symbols ($24$ addition signs and $3$ ones) will be addition signs or, alternatively, choose which $3$ of the $27$ symbols will be ones.
A: For $x$ we have $25$ different choices. Then for $y$ we have to choose from $x$ to $25$ and for $z$ we must choose between $y$ and $25$. So the total number of such triples are
$$\sum_{x=1}^{25} \sum_{y=x}^{25} \sum_{z=y}^{25} 1.$$
A: Given $y$ there are $y$ possibilities for $x$. Given $z$ there are $z$ possibilities for $y$ and hence $1+2+\dots+z=\frac{1}{2}z(z+1)$ for $(x,y)$. Hence the answer is $\sum_1^{25}\frac{1}{2}z(z+1)=\frac{1}{6}25\cdot26\cdot27=2925$
