Inclusion Exclusion principle and programming applications
Sum Rule – If a task can be done in one of ways or one of ways, where none of the set of ways is the same as any of the set of ways, then there are ways to do the task.
The sumrule mentioned above states that if there are multiple sets of ways of doing a task, there shouldn’t be any way that is common between two sets of ways because if there is, it would be counted twice and the enumeration would be wrong.
The principle of inclusionexclusion says that in order to count only unique ways of doing a task, we must add the number of ways to do it in one way and the number of ways to do it in another and then subtract the number of ways to do the task that are common to both sets of ways.
The principle of inclusionexclusion is also known as the subtraction principle. For two sets of ways and , the enumeration would like
Below are some examples to explain the application of inclusionexclusion principle:

Example 1: How many binary strings of length 8 either start with a ‘1’ bit or end with two bits ’00’?
Solution: If the string starts with one, there are 7 characters left which can be filled in ways.
If the string ends with ’00’ then 6 characters can be filled in ways.
Now if we add the above sets of ways and conclude that it is the final answer, then it would be wrong. This is because there are strings with start with ‘1’ and end with ’00’ both, and since they satisfy both criteria they are counted twice.
So we need to subtract such strings to get a correct count.
Strings that start with ‘1’ and end with ’00’ have five characters that can be filled in ways.
So by the inclusionexclusion principle we get
Total strings = 128 + 64 – 32 = 160 
Example 2: How many numbers between 1 and 1000, including both, are divisible by 3 or 4?
Solution: Number of numbers divisible by 3 = =.
Number of numbers divisible by 4 = = .
Number of numbers divisible by 3 and 4 = = .
Therefore, number of numbers divisible by 3 or 4 = = 333 + 250 – 83 = 500
Implementation
Problem 1: How many numbers between 1 and 1000, including both, are divisible by 3 or 4?
The Approach will be the one discussed above, we add the number of numbers that are divisible by 3 and 4 and subtract the numbers which are divisible by 12.
C++
// CPP program to count the // number of numbers between // 1 and 1000, including both, // that are divisible by 3 or 4 #include <bits/stdc++.h> using namespace std; // function to count the divisors int countDivisors( int N, int a, int b) { // Counts of numbers // divisible by a and b int count1 = N / a; int count2 = N / b; // inclusionexclusion // principle applied int count3 = (N / (a * b)); return count1 + count2  count3; } // Driver Code int main() { int N = 1000, a = 3, b = 4; cout << countDivisors(N, a, b); return 0; } 
Java
// Java program to count the // number of numbers between // 1 and 1000, including both, // that are divisible by 3 or 4 import java.io.*; class GFG { // function to count the divisors public static int countDivisors( int N, int a, int b) { // Counts of numbers // divisible by a and b int count1 = N / a; int count2 = N / b; // inclusionexclusion // principle applied int count3 = (N / (a * b)); return count1 + count2  count3; } // Driver Code public static void main (String[] args) { int N = 1000 , a = 3 , b = 4 ; System.out.println (countDivisors(N, a, b)); } } // This code is contributed by m_kit 
Python3
# Python3 program to count the # number of numbers between # 1 and 1000, including both, # that are divisible by 3 or 4 # function to count the divisors def countDivisors(N, a, b): # Counts of numbers # divisible by a and b count1 = N / / a count2 = N / / b # inclusionexclusion # principle applied count3 = (N / / (a * b)) return count1 + count2  count3 # Driver Code N = 1000 a = 3 b = 4 print (countDivisors(N, a, b)) # This code is contributed by shubhamsingh10 
C#
// C# program to count the // number of numbers between // 1 and 1000, including both, // that are divisible by 3 or 4 using System; class GFG { // function to count // the divisors public static int countDivisors( int N, int a, int b) { // Counts of numbers // divisible by a and b int count1 = N / a; int count2 = N / b; // inclusionexclusion // principle applied int count3 = (N / (a * b)); return count1 + count2  count3; } // Driver Code static public void Main () { int N = 1000, a = 3, b = 4; Console.WriteLine(countDivisors(N, a, b)); } } // This code is contributed by aj_36 
PHP
<?php // PHP program to count the // number of numbers between // 1 and 1000, including both, // that are divisible by 3 or 4 // function to count the divisors function countDivisors( $N , $a , $b ) { // Counts of numbers // divisible by a and b $count1 = $N / $a ; $count2 = $N / $b ; // inclusionexclusion // principle applied $count3 = ( $N / ( $a * $b )); return $count1 + $count2  $count3 ; } // Driver Code $N = 1000; $a = 3; $b = 4; echo countDivisors( $N , $a , $b ); // This code is contributed by aj_36 ?> 
Output :
500
.
Problem 2: Given N prime numbers and a number M, find out how many numbers from 1 to M are divisible by any of the N given prime numbers.
Examples :
Input: N numbers = {2, 3, 5, 7} M = 100 Output: 78 Input: N numbers = {2, 5, 7, 11} M = 200 Output: 137
The approach for this problem will be to generate all the possible combinations of numbers using N prime numbers using power set in 2^{N}. For each of the given prime numbers P_{i} among N, it has M/P_{i} multiples. Suppose M=10, and we are given with 3 prime numbers(2, 3, 5), then the total count of multiples when we do 10/2 + 10/3 + 10/5 is 11. Since we are counting 6 and 10 twice, the count of multiples in range 1M comes 11. Using inclusionexclusion principle, we can get the correct number of multiples. The inclusionexclusion principle for three terms can be described as:
Similarly, for every N numbers, we can easily find the total number of multiples in range 1 to M by applying the formula for an intersection of N numbers. The numbers that are formed by multiplication of an odd number of prime numbers will be added and the numbers formed by multiplication of even numbers will thus be subtracted to get the total number of multiples in the range 1 to M.
Using power set we can easily get all the combination of numbers formed by the given prime numbers. To know if the number is formed by multiplication of odd or even numbers, simply count the number of set bits in all the possible combinations (11<<N).
Using power sets and adding the numbers created by combinations of odd and even prime numbers we get 123 and 45 respectively. Using inclusionexclusion principle we get the number of numbers in range 1M that is divided by any one of N prime numbers is (odd combinationseven combinations) = (12345) = 78.
Below is the implementation of the above idea:
C++
// CPP program to count the // number of numbers in range // 1M that are divisible by // given N prime numbers #include <bits/stdc++.h> using namespace std; // function to count the number // of numbers in range 1M that // are divisible by given N // prime numbers int count( int a[], int m, int n) { int odd = 0, even = 0; int counter, i, j, p = 1; int pow_set_size = (1 << n); // Run from counter 000..0 to 111..1 for (counter = 1; counter < pow_set_size; counter++) { p = 1; for (j = 0; j < n; j++) { // Check if jth bit in the // counter is set If set // then pront jth element from set if (counter & (1 << j)) { p *= a[j]; } } // if set bits is odd, then add to // the number of multiples if (__builtin_popcount(counter) & 1) odd += ( m / p ); else even += ( m / p ); } return odd  even; } // Driver Code int main() { int a[] = { 2, 3, 5, 7 }; int m = 100; int n = sizeof (a) / sizeof (a[0]); cout << count(a, m, n); return 0; } 
Python3
# Python3 program to count the # number of numbers in range # 1M that are divisible by # given N prime numbers def bitsoncount(x): return bin (x).count( '1' ) # function to count the number # of numbers in range 1M that # are divisible by given N # prime numbers def count(a, m, n): odd = 0 even = 0 p = 1 pow_set_size = 1 << n # Run from counter 000..0 to 111..1 for counter in range ( 1 ,pow_set_size): p = 1 for j in range (n): # Check if jth bit in the # counter is set If set # then pront jth element from set if (counter & ( 1 << j)): p * = a[j] # if set bits is odd, then add to # the number of multiples if (bitsoncount(counter) & 1 ): odd + = ( m / / p ) else : even + = ( m / / p ) return odd  even # Driver Code a = [ 2 , 3 , 5 , 7 ] m = 100 n = len (a) print (count(a, m, n)) # This code is contributed by shivanisinghss2110 
Output:
78
Time Complexity : O(2^{N}*N)
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