Count of All Permutations on n Objects/Examples/Even Integers from 1, 2, 3, 4

Example of Count of All Permutations on $n$ Objects

Let $N$ be the number of even integers which can be made using one or more of the digits $1$, $2$, $3$ and $4$ no more than once each.

Then:

$N = 32$

From Count of All Permutations on $n$ Objects, the total number of integers which can be made using the digits $1$, $2$, $3$ and $4$ is given by:

$\ds 4! \sum_{k \mathop = 0}^3 \dfrac 1 {k!}$

$=$

$\ds 24 \paren {\dfrac 1 {0!} + \dfrac 1 {1!} + \dfrac 1 {2!} + \dfrac 1 {3!} }$

$\ds $

$=$

$\ds \dfrac {24} 1 + \dfrac {24} 1 + \dfrac {24} 2 + \dfrac {24} 6$

$\ds $

$=$

$\ds 24 + 24 + 12 + 4$

$\ds $

$=$

$\ds 64$

By symmetry, exactly half of these integers end in $2$ or $4$.

The others end in $1$ or $3$.

Hence:

$N = \dfrac {64} 2 = 32$

$\blacksquare$