Combinatorics

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Binomial Coefficients

The binomial coefficient $\binom{n}{k}$ (pronounced as "$n$ choose $k$" or sometimes written as ${}_nC_k$) represents the number of ways to choose a subset of $k$ elements from a set of $n$ elements. For example, $\binom{4}{2} = 6$, because the set $\{1,2,3,4\}$ has $6$ subsets of $2$ elements:

There are two ways to calculate binomial coefficients:

Method 1: Pascal's Triangle (Dynamic Programming) - $\mathcal{O}(n^2)$

Binomial coefficients can be recursively calculated as follows:

The intuition behind this is to fix an element $x$ in the set and choose $k − 1$ elements from $n − 1$ elements if $x$ is included in the set or choose $k$ elements from $n − 1$ elements, otherwise.

The base cases for the recursion are:

because there is always exactly one way to construct an empty subset and a subset that contains all the elements.

This recursive formula is commonly known as Pascal's Triangle.

A naive implementation of this would use a recursive formula, like below:

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/** Computes nCk mod p using naive recursion */
int binomial(int n, int k, int p) {
	if (k == 0 || k == n) { return 1; }
	return (binomial(n - 1, k - 1, p) + binomial(n - 1, k, p)) % p;
}

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/** Computes nCk mod p using naive recursion */
public static int binomial(int n, int k, int p) {
	if (k == 0 || k == n) { return 1; }
	return (binomial(n - 1, k - 1, p) + binomial(n - 1, k, p)) % p;
}

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def binomial(n: int, k: int, p: int) -> int:
	"""Computes nCk mod p using naive recursion"""
	if k == 0 or k == n:
		return 1
	return (binomial(n - 1, k - 1, p) + binomial(n - 1, k, p)) % p

Additionally, we can optimize this from $\mathcal{O}(2^n)$ to $\mathcal{O}(n^2)$ using dynamic programming (DP) by caching the values of smaller binomials to prevent recalculating the same values over and over again. The code below shows a bottom-up implementation of this.

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/** Computes nCk mod p using dynamic programming */
int binomial(int n, int k, int p) {
	// dp[i][j] stores iCj
	vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));

	// base cases described above
	for (int i = 0; i <= n; i++) {
		/*
		 * i choose 0 is always 1 since there is exactly one way
		 * to choose 0 elements from a set of i elements

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/** Computes nCk mod p using dynamic programming */
public static int binomial(int n, int k, int p) {
	// dp[i][j] stores iCj
	int[][] dp = new int[n + 1][k + 1];

	// base cases described above
	for (int i = 0; i <= n; i++) {
		/*
		 * i choose 0 is always 1 since there is exactly one way
		 * to choose 0 elements from a set of i elements

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def binomial(n: int, k, p):
	"""Computes nCk mod p using dynamic programming"""
	# dp[i][j] stores iCj
	dp = [[0] * (k + 1) for _ in range(n + 1)]

	# base cases described above
	for i in range(n + 1):
		"""
		i choose 0 is always 1 since there is exactly one way
		to choose 0 elements from a set of i elements

Method 2: Factorial Definition (Modular Inverses) - $\mathcal{O}(n + \log MOD)$

Define $n!$ as $n \times (n - 1) \times (n - 2) \times \ldots 1$. $n!$ represents the number of permutations of a set of $n$ elements. See this AoPS Article for more details.

Another way to calculate binomial coefficients is as follows:

Recall that $\binom{n}{k}$ also represents the number of ways to choose $k$ elements from a set of $n$ elements. One strategy to get all such combinations is to go through all possible permutations of the $n$ elements, and only pick the first $k$ elements out of each permutation. There are $n!$ ways to do so. However, note the the order of the elements inside and outside the subset does not matter, so the result is divided by $k!$ and $(n − k)!$.

Since these binomial coefficients are large, problems typically require us to output the answer modulo a large prime $p$ such as $10^9 + 7$. Fortunately, we can use modular inverses to divide $n!$ by $k!$ and $(n - k)!$ modulo $p$ for any prime $p$. Computing inverse factorials online can be very time costly. Instead, we can precompute all factorials in $\mathcal{O}(n)$ time and inverse factorials in $\mathcal{O}(n + \log MOD)$. First, we compute the modular inverse of the largest factorial using binary exponentiation. For the rest, we use the fact that $(n!)^{-1} \equiv (n!)^{-1}\times (n+1)^{-1} \times (n+1) \equiv ((n+1)!)^{-1}\times (n+1)$. See the code below for the implementation.

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const int MAXN = 1e6;

long long fac[MAXN + 1];
long long inv[MAXN + 1];

/** Computes x^n modulo m in O(log p) time. */
long long exp(long long x, long long n, long long m) {
	x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow
	long long res = 1;
	while (n > 0) {

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import java.util.*;

public class BinomialCoefficients {
	private static final int MAXN = (int)1e6;
	private static long[] fac = new long[MAXN + 1];
	private static long[] inv = new long[MAXN + 1];

	/** Computes x^n modulo m in O(log p) time. */
	private static long exp(long x, long n, long m) {
		x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow

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MAXN = 10**6

fac = [0] * (MAXN + 1)
inv = [0] * (MAXN + 1)


def exp(x: int, n: int, m: int) -> int:
	"""Computes x^n modulo m in O(log p) time."""
	x %= m  # note: m * m must be less than 2^63 to avoid ll overflow
	res = 1

Solution - Binomial Coefficients

The first method for calculating binomial factorials is too slow for this problem since the constraints on $a$ and $b$ are $(1 \leq b \leq a \leq 10^6)$ (recall that the first implementation runs in $\mathcal{O}(n^2)$ time complexity). However, we can use the second method to answer each of the $n$ queries in constant time by precomputing factorials and their modular inverses.

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#include <iostream>
using namespace std;
using ll = long long;

const int MAXN = 1e6;
const int MOD = 1e9 + 7;

ll fac[MAXN + 1];
ll inv[MAXN + 1];

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import java.io.*;
import java.util.*;

public class BinomialCoefficients {
	private static final int MAXN = (int)1e6;
	private static final int MOD = (int)1e9 + 7;
	private static long[] fac = new long[MAXN + 1];
	private static long[] inv = new long[MAXN + 1];

	public static void main(String[] args) {

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MAXN = 10**6
MOD = 10**9 + 7

fac = [0] * (MAXN + 1)
inv = [0] * (MAXN + 1)


 Code Snippet: Counting Functions (Click to expand)

Derangements

The number of derangements of $n$ numbers, expressed as $!n$, is the number of permutations such that no element appears in its original position. Informally, it is the number of ways $n$ hats can be returned to $n$ people such that no person recieves their own hat.

Method 1: Principle of Inclusion-Exclusion

Suppose we had events $E_1, E_2, \dots, E_n$, where event $E_i$ corresponds to person $i$ recieving their own hat. We would like to calculate $n! - \lvert E_1 \cup E_2 \cup \dots \cup E_n \rvert$.

We subtract from $n!$ the number of ways for each event to occur; that is, consider the quantity $n! - \lvert E_1 \rvert - \lvert E_2 \rvert - \dots - \lvert E_n \rvert$. This undercounts, as we are subtracting cases where more than one event occurs too many times. Specifically, for a permutation where at least two events occur, we undercount by one. Thus, add back the number of ways for two events to occur. We can continue this process for every size of subsets of indices. The expression is now of the form:

For a set size of $k$, the number of permutations with at least $k$ indicies can be computed by choosing a set of size $k$ that are fixed, and permuting the other indices. In mathematical terms:

Thus, the problem now becomes computing

which can be done in linear time.

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#include <bits/stdc++.h>

// https://atcoder.github.io/ac-library/document_en/modint.html
// (included in grader for this problem)
#include <atcoder/modint>
using mint = atcoder::modint;

using namespace std;

int main() {

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import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int N = scanner.nextInt();
		int M = scanner.nextInt();

		long c = 1;
		for (int i = 1; i <= N; i++) {

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N, M = map(int, input().split())
c = 1
for i in range(1, N + 1):
	c = (c * i) + (-1 if i % 2 == 1 else 1)
	c %= M
	c += M
	c %= M
	print(c, end=" ")

print()

Method 2: Dynamic Programming

Suppose person 1 recieved person $i$'s hat. There are two cases:

1. If person $i$ recieves person 1's hat, then the problem is reduced to a subproblem of size $n - 2$. There are $n - 1$ possibilities for $i$ in this case, so we add to the current answer $(n - 1)\cdot !(n - 2)$.
2. If person $i$ does not recieve person 1's hat, then we can reassign person 1's hat to be person $i$'s hat (if they recieved person 1's hat, then this would become first case). Thus, this becomes a subproblems with size $n - 1$, are there $n - 1$ ways to choose $i$.

Thus, we have

which can be computed in linear time with a simple DP. The base cases are that $!0 = 1$ and $!1 = 0$.

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#include <bits/stdc++.h>

// https://atcoder.github.io/ac-library/document_en/modint.html (included in
// grader)
#include <atcoder/modint>
using mint = atcoder::modint;

using namespace std;

int main() {

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import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();

		int a = 1;
		int b = 0;

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N, M = map(int, input().split())
a, b = 1, 0

print(0, end=" ")
for i in range(2, N + 1):
	c = (i - 1) * (a + b) % M
	print(c, end=" ")
	a, b = b, c

print()

Problems