Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding - Page 875: 29

$C(4, 3)=4$ The list of possible combinations: $123,124,134, 234$

If we want to choose $r$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $C(n, r)=\frac{n!}{(n-r)!r!}$ ways. Hence, $C(4, 3)=\dfrac{4!}{(4-3)!3!}=\dfrac{4!}{1!3!}=4$ The list of all possible combinations: $123,124,134, 234$