A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
Here is another way.
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Probability is favorable to possible.
What is favorable?
{1,2,3} All premutations of this set = 6
{2,3,4} All premutations of this set = 6
.....
{18,19,20} All premutations of this set = 6
In total we have,
18 Sets
Thus,
$\displaystyle 18\cdot 6=108$
The possible outcomes is the number of ways of selecting 3 out of twenty.
Which is,
$\displaystyle _{20}C_3=1140 $
Thus,
$\displaystyle \frac{108}{1140}$
There are 18 subsets of three consecutive integers.
There are $\displaystyle _{20}C_3=1140 $ total subsets three integers.
Thus, $\displaystyle \frac{18}{1140} = 0.01579$
The order does not come to this.
If we do use order then we would have $\displaystyle \frac{108}{(20)(19)(18)} = 0.01579$