Probability - consecutive numbers Question:

Three numbers are selected out of the first 30 natural numbers. What
  is the probability that none of them are consecutive?

I know that the total possibilities will be $^{30}C_3$
However, I'm not sure how to calculate the number of combinations that have consecutive numbers?
 A: A bijective proof: 
The compression map 
$(x,y,z)\mapsto (x,y-1,z-2)$
is a bijection between increasing sequences of three numbers 
drawn from $\{1,2,\dots, 30\}$ without consecutive values and 
increasing sequences of three numbers drawn from $\{1,2,\dots, 28\}$.
Hence, there are ${28\choose 3}=3276$ such sets.   
A: The number of ways of picking numbers so that two but not three are consecutive is given by:
$$N(2)=2(28-1) + (29-2)(28-2) = 756$$
since there are:


*

*two ways of choosing consecutive numbers at very beginning or end of $\{1,2,3,\ldots,28,29,30\}$ and then $28-1$ possible choices of a third non-adjacent number

*$29-2$ ways of choosing consecutive numbers in the middle of of $\{1,2,3,\ldots,28,29,30\}$ and then $28-2$ possible choices of a third non-adjacent number


while the number of ways of choosing three consecutive numbers is given by:
$$N(3)=30-2 = 28$$
So the number of non-consecutive choices is:
$$\binom{30}{3}-N(2)-N(3)=4060-756-28=3276$$
and the probability is:
$$P=\frac{3276}{4060}\approx0.807$$
A: We can use the Inclusion-Exclusion Principle to determine how many three-element subsets of the set $\{1, 2, 3, \ldots, 30\}$ do not contain consecutive numbers.
As you found, there are $\binom{30}{3}$ three-element subsets.  From these, we must remove those with at least two consecutive numbers.  There are $29$ pairs of consecutive numbers since the smaller number in any pair of consecutive numbers in the set $\{1, 2, 3, \ldots, 30\}$ cannot exceed $29$.  We can then pick the third element in the set in $28$ ways.  Thus, there are $29 \cdot 28 = 812$ three-element subsets that contain two consecutive numbers.  However, we have counted those that contain three consecutive numbers twice, once when we counted the smallest two numbers and once when we counted the largest two numbers in the subset.  There are $28$ subsets consisting of three consecutive integers since the smallest number in the subset cannot be $29$ or $30$.  Therefore, the number of three-element subsets of the set $\{1, 2, 3, \ldots, 30\}$ that do not contain consecutive integers is 
$$\binom{30}{3} - 29 \cdot 28 + 28 = 4060 - 812 + 28 = 3276$$
A: A less technical explanation.
Consider 30 unlabelled counters (C). Take away 3, now 27 are left, and there are 28 gaps (including ends) where we can insert back the 3 to comply with non-consecutive stipulation. Now serially number the counters.
_C_C_C_C_C_C_C_ ........ C_C_C_C_C_C_C_C_C_C_
$$\text{thus there can be}\;\;{28\choose3} = 3276\;\; \text{such subsets}$$
$$\text{And the probability} = \frac{28\choose 3}{30\choose 3} = \frac{117}{145}$$
A: Consider any 3 consecutive numbers as a single set. Thus, you'll get as many as $28$ possible cases of picking such a set out of first 30 natural numbers.
Since, the order of picking the $3$ numbers has not yet been accounted for, it'd be $3!$. Hence, the final answer would be: $\binom{30}{3} - 3! \cdot 28 = 4060 - 168 = 3892$
