91Ó°ÊÓ

Let$A$be the event that the number-$1$the horse is among the top three finishers, and let$B$be the event that the number-$2$ horse comes in second.

Using only mathematical context we have:

Outcome space$S$:

$S=\left\{\left(x_1,x_2,…x_6\right):x_i∈\{1,2,…6\},âˆ\P Äi\right\}$

$S$has all the ordered permutations of numbers $1-6$(representing horses).

$\left|S\right|=6!$- the number of elements it $S$is$6!$.

Event $A$- the number$1$is eitherlocalid="1649427017351" $x_1,x_2$-orlocalid="1649427028219" $x_3$
.

$A=\left\{\left(1,x_2,…x_6\right),\left(x_1,1,…x_6\right),\left(x_1,x_2,1,…x_6\right):x_i∈\{2,…6\},âˆ\P Äi\right\}$

There are $5!$permutations of the other elements of the vector $(2,3,4,5$&$6)$and $1$can be the first the second, or the third.

By the multiplication principle of counting - $\left|A\right|=3·5!$

Event$B$- the number $2$is$x_2$.

$B=\left\{\left(x_1,2,…x_6\right):x_i∈\{1,3,…6\},âˆ\P Äi\right\}$

There are $5!$permutations of the other elements of the vector$(1,3,4,5$$\&6)$.$\left|B\right|=5!$

An event $Aâˆ\copyright B$where number one is in the first three and the number two is second

$Aâˆ\copyright B=\left\{\left(1,2,…x_6\right),\left(x_1,2,1,…x_6\right):x_i∈\{3,…6\},âˆ\P Äi\right\}$

The remaining numbers$-3,4,5$&$6$can be permuted in $4!$ways. And $1,2$can either be the first and the second number or the third and the second number.

$|Aâˆ\copyright B|=2·4!$

From one of the previous exercises we know:

$|A∪B|=\left|A\right|+\left|B\right|-|Aâˆ\copyright B|$

The other notation for the number of elements in a set is$N\left(A\right)$.

Finally:

$|A∪B|=3·5!+5!-2·4!=18·4!=432$