A probability problem by Kislay Raj

The number of ways to select three numbers from twenty without regard for order is ${20 \choose 3} = 1140$

The number of ways to choose three numbers from the first positive twenty integers which have exactly two consecutive integers (i.e. explicitly avoiding the case with three consecutive integers to avoid double counting) without regard for order is:

$...$

$17*2+16*17=306$

The number of ways to choose three numbers from the first positive twenty integers which have exactly three consecutive integers without regard for order is:

$(1,2,3)$

$(2,3,4)$

$(3,4,5)$

$...$

$(18,19,20)$

$=18$

$1140-306-18=816$

The number of ways to select 3 numbers is $\left( \begin{matrix} 20 \\ 3 \end{matrix} \right) = 1140$

CASE 1 - The $3$ numbers that have been selected are all consecutive.

Thus we would get $\_18\_$ sets of $3$ consecutive numbers - $\left( 1,2,3 \right) ,\left( 2,3,4 \right) ,\left( 3,4,5 \right) ......\left( 18,19,20 \right)$

CASE 2 - Any $2$ of the $3$ numbers are consecutive.

Here

PART 1 - We have the sets $\left( 1,2,x \right)$ and $\left( y,19,20 \right)$ $x$ and $y$ each can have 17 values (since $x$ cannot be $3$ and $y$ cannot be $18$ ) Thus we get $17 x 2=\_34\_$

PART 2 - We have sets $\left( x,2,3,y \right) ............\left( a,18,19,b \right)$

( Two variables because either one can make the set a set of $3$

Here $x$ and $y$ cannot have the values $1$ and $2$ , so we get $16$ values for the set . Similarly we have 16 values for all the remaining sets up to $\left( a,18,19,b \right)$ . Thus we get $16 x 18=\_288\_$

But, here we have OVER COUNTED #Principle of inclusion and exclusion and need to reduce a $\_16\_$

Thus we finally get $1140-18-34-288+16=816$

Let a=1. The smallest possible value for b=3. If b=3, c ranges from 5 to 20 so 16 values. If b=4 then c ranges from 6 to 20 so 15 values and so on and so forth........ The largest value of b=18 and even then c=20 so 1 case.

For a=1, we have 16+15 + ...+1 =16C1 + 15C1 + ...+ 1C1 = 17C2 by Hockey Stick Identity

For a=2, we have 15+14+....+1 = 15C1 + 14C1 + ...+ 1C1 = 16C2. Note that we start with 15 instead of 16 here as since a=2, b must be at least 4 so one less value for c.

For a=3, you can roughly guess we have 14C1+13C1 + ....+1C1 = 15C2.

Do you see a pattern? For subsequent values of a we have nC2 where n decreases by 1 each time.

Total number of cases = 17C2 + 16C2 + 15C2 + 14C2 + ...+3C2 + 2C2 = 18C3 = $\boxed{816}$ .