How many $5$ element sets can be made? 
Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$.

I think the opposite would be to find, the number of sets with no consecutive numbers.
There are: $\binom{14}{5} = 2002$ total sets. 
$M = \{a, b, c, d, e\}$
Now I am confused, how do I find sets with no consecutive numbers?
 A: What you want to get is a line formed by $9$ balls and $5$ bars, not being two bars consecutive. The bars represent the number you choose, The balls, the number you don't.
So first, we must put a ball between each pair of bars. We have no choice: this can be done in $1$ way.
Now we can distribute freely the remainding balls. This is like putting $5$ balls in $6$ urns, a bars and stars problem; thus, the number of subsets with five no consecutive elements is 
$$\binom{10}5$$
A: There are $\binom{n-k+1}{k}$ subsets of $\{1,2,3\dots n\}$ with $k$ elements and no two consecutive.
There is a bijective proof. Given a subset of $\{1,2,3,\dots n-k+1\}$ with elements $a_1,a_2\dots a_k$ in increasing order. send them to the following elements:
$a_1\mapsto a_1$
$a_2\mapsto a_2+1$
$a_3\mapsto a_3+2$
$\dots$ 
$a_k\mapsto a_k+k-1$
The new terms $a_1,a_2+1,\dots ,a_k+k-1$ form a subset of $k$ elements of $\{1,2,3\dots n\}$ with no consecutive elements. This function clearly has an inverse, we can give it explicitly:
$a_1\mapsto a_1$
$a_2\mapsto a_2-1$
$a_3\mapsto a_3-2$
$\dots$ 
$a_k\mapsto a_k-k+1$
Since it has an inverse it is bijective and so the proof is established.
Hence there are $\binom{14}{5}$ total and $\binom{10}{5}$ with no consecutive terms. So the final answer is $\binom{14}{5}-\binom{10}{5}=1750$. Of course you only want the three last digits, which are $750$.
A: You can use $H^n_r$ here, but take out 5 first, for the abcde, then 4, to insert between a,b; b,c; c,d; d,e.
Others can be 'inserted' freely in the 6 'boxes' separated by the five numbers.
Number of sets with no consecutive numbers $=H_{14-5-4}^6=H_5^6=C^{10}_5$
A: There are $\binom{10}{5}$ ways to pick out $5$ distinct numbers out of $\left\{ 1,\dots,10\right\} $.
The result $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ corresponds with $a_{1}<a_{2}+1<a_{3}+2<a_{4}+3<a_{5}+4$
wich can be seen as a pick out of $5$ distinct and non-consecutive numbers out
of $\left\{ 1,\dots,14\right\} $.
A: We wish to calculate the number of subsets of five elements of the set $\{1, 2, 3, \ldots, 14\}$ in which no two of the numbers are consecutive.
Suppose the five non-consecutive numbers selected from the set $\{1, 2, 3, \ldots, 14\}$ are $x_1, x_2, x_3, x_4, x_5$, where $x_1 < x_2 < x_3 < x_4 < x_5$.  Let 
\begin{align*}
y_1 & = x_1\\
y_2 & = x_2 - x_1\\
y_3 & = x_3 - x_2\\
y_4 & = x_4 - x_3\\
y_5 & = x_5 - x_4\\
y_6 & = 14 - x_5 
\end{align*}
Since $y_1$ is the smallest number in the subset, $y_1 \geq 1$.  Since no two elements of the subset $\{x_1, x_2, x_3, x_4, x_5\}$ are consecutive, $y_2, y_3, y_4, y_5 \geq 2$. The number $y_6$ represents the difference between $14$ and the largest number in the subset, so $y_6 \geq 0$.  Moreover,
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 14 \tag{1}$$
Let 
\begin{align*}
z_1 & = y_1 - 1\\
z_2 & = y_2 - 2\\
z_3 & = y_3 - 2\\
z_4 & = y_4 - 2\\
z_5 & = y_5 - 2\\
z_6 & = y_6
\end{align*}
Substitution into equation 1 yields
\begin{align*}
z_1 + 1 + z_2 + 2 + z_3 + 2 + z_4 + 2 + z_5 + 2 + z_6 & = 14\\
z_1 + z_2 + z_3 + z_4 + z_5 + z_6 & = 5 \tag{2}
\end{align*}
which is an equation in the nonnegative integers.  The number of solutions of equation 2 is the number of ways five addition signs can be inserted into a row of five ones, which is 
$$\binom{5 + 5}{5} = \binom{10}{5}$$
Hence, the number of subsets in which no two of the numbers are consecutive is $\binom{10}{5}$. 
