Factorials and Integer Partitions

math
combinatorics
Author

sun123zxy

Published

June 15, 2025

Factorials are commonly seen in combinatorics, and today we talk about some variants of them related to integer partitions. Recall that if one wish to count the number of ways to choose \(k\) objects from \(n\) distinct objects, we have binomial coefficients defined as \[ \binom n k := \frac{n!}{k!(n-k)!} \] A slightly more general case is when we want to partition \(n\) distinct objects into several labeled sets, with the size of each set is identified by an integer partition \((\lambda_1,\lambda_2,\dots) = \lambda \vdash n\). In such circumstance, the multinomial coefficients, \[ \binom{n}{\lambda} := \frac{n!}{\lambda!} := \frac{n!}{\prod_i \lambda_i!} \] counts the number of ways to do so. So here comes the first variant \[ \lambda! := \prod_i \lambda_i! \]

Remark (OGF magic)

Remark (OGF magic). A fancy way to find this number:

\[ \frac{1}{1 - e_1} = \sum_{n \in \mathbb N} e_1^n = \sum_{n \in \mathbb N} \sum_{\lambda \vdash n} \binom{n}{\lambda} m_\lambda \]

What if these sets are indistinguishable? Some solutions may be counted multiple times. For example, in the \(\lambda = (n,n) \vdash 2n\) case, it does not matter which set is the first or the second.

In this case, we need to divide by \(2!\) to cancel out the order of the two sets. In general, sets with the same size are indistinguishable, so we need to divide by the product of factorials of the multiplicities of each set size. This leads us to the second variant.

Recall that another way to write an integer partition is to apply the multiplicity notation, that is, we count the number of occurrences \(c_i\) of the number \(i\) in the sequence \((\lambda_1,\lambda_2,\dots)\) to write \(\lambda = 1^{c_1} 2^{c_2} 3^{c_3} \dots\).Now we may define \[ \lambda^! := \prod_i c_i! \] and the number of ways to partition \(n\) distinct objects into indistinguishable sets of sizes \(\lambda\) is now given by \[ \frac{n!}{\lambda! \cdot \lambda^!} \]

Remark (EGF magic)

Remark (EGF magic). A fancy way to find this number:

\[ [t^c] \exp \left( \sum_{k \geq 1} \frac 1 {k!} t_k \right) = \frac{1}{\lambda! \cdot \lambda^!} \]

We provide some other examples of \(\lambda^!\). First, the number of permuations of cycle type \(\lambda\) is given by \[ \frac{n!}{(\prod_i \lambda_i) \cdot \lambda^!} \] whose denominator is known as the coefficients of the \(n\)-th cycle indicator \(Z_n\). Second, when computing the chromatic symmetric function of a graph by hand, it’s often the case that we first partition the vertices into groups, and then assign colors to these groups. Say these groups are of sizes \(\lambda\), then upon assignment, \(\lambda^! m_\lambda\) is contributed to the chromatic symmetric function.