Pick any six numbers from 1 to 49. What is the probability that none of them are consecutive? Pick any six numbers from 1 to 49. What is the probability that none of them are consecutive ?
I think it should be $\displaystyle \frac{^{49}C_1\cdot^{47}C_1\cdot^{45}C_1\cdot^{43}C_1\cdot^{41}C_1\cdot^{39}C_1}{^{49}C_6}$
Is this correct ?
 A: There are $\binom{49}{6}$ equally likely ways to pick $6$ numbers. That will be the denominator. The numerator is the number of "favourables" which we now count.
Write down a sequence of $43$ circles $\circ$, to represent numbers not chosen, like this:
$$\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad\circ\quad$$
(we stopped short of $43$). These determine $44$ gaps ($2$ of which are at the ends) from which we choose $6$ to slip an x into to represent a chosen position. 
There are $\binom{44}{6}$ ways to do this.
Remark: The denominator you had produced is correct. Your numerator does not correctly count the favourables.
A: The prohibition on consecutive numbers means the constraints are really 
$$1 \le a_1$$ 
$$a_1+1 \lt a_2$$ 
$$a_2+1 \lt a_3$$ 
$$a_3+1 \lt a_4$$ 
$$a_4+1 \lt a_5$$ 
$$a_5+1 \lt a_6$$ 
$$a_6 \le 49$$
Rewrite these so that the right hand side of each line is the same as the left hand side of the next line as 
$$1 \le a_1$$ 
$$a_1 \lt a_2-1$$ 
$$a_2-1 \lt a_3-2$$ 
$$a_3-2 \lt a_4-3$$ 
$$a_4-3 \lt a_5-4$$ 
$$a_5-4 \lt a_6-5$$ 
$$a_6-5 \le 44$$ 
Now you can now combine these as a single line 
$$1 \le a_1 < a_2 — 1 < a_3 — 2 < a_4 — 3 < a_5 — 4 < a_6 — 5 \le 44$$  
Having done that, you could let $b_1=a_1$, $b_2=a_2-1$, $b_3=a_3-2$, $b_4=a_4-3$, $b_5=a_5-4$, and $b_6=a_6-5$ so $$1 \le b_1 < b_2  < b_3 < b_4  < b_5 < b_6 \le 44$$ and so the number of integer solutions for the $b_i$ is obviously the same as the number of ways of choosing $6$ distinct integers from $44$, i.e. $\displaystyle ^{44}C_6$. 
Having chosen the $b_i$ in order from smallest to largest, you can get back to the $a_i$ by adding $0$ to the smallest, $1$ to the second, $2$ to the third, $3$ to the fourth, $4$ to the fifth and $5$ to the sixth(largest). It is obvious that this will give six numbers with at least one gap between them.
Adapted from Choice Problem: choose 5 days in a month, consecutive days are forbidden
