Here a direct bijective proof is not possible: because the right-hand side of the identity is a fraction, there is no set "obviously" counted by it (it even takes some thought to see that the denominator always evenly divides the numerator). However its numerator counts the Cartesian product of "k" finite sets of sizes "n", , ..., , while its denominator counts the permutations of a "k"-element set (the set most obviously counted by the denominator would be another Cartesian product "k" finite sets; if desired one could map permutations to that set by an explicit bijection). Now take "S" to be the set of sequences of "k" elements selected from our "n"-element set without repetition. On one hand, there is an easy bijection of "S" with the Cartesian product corresponding to the numerator formula_3, and on the other hand there is a bijection from the set "C" of pairs of a "k"-combination and a permutation "σ" of "k" to "S", by taking the elements of "C" in increasing order, and then permuting this sequence by "σ" to obtain an element of "S". The two ways of counting give the equation