Find the Number of Possible Ways for an Event

Find the Number of Possible Ways for an Event - Problem

Music Festival Challenge: Stage Assignments and Band Scoring

Imagine you're organizing a massive music festival! You have n talented performers who need to be assigned to x different stages. Here's how it works:

🎭 Stage Assignment: Each performer must be assigned to exactly one of the x stages
🎵 Band Formation: All performers on the same stage form a band (some stages might be empty)
🏆 Scoring: After performances, each non-empty band receives a score from 1 to y

Your mission: Calculate the total number of different ways this entire event can unfold!

Two events are considered different if:
• Any performer is assigned to a different stage, OR
• Any band receives a different score

Since the answer can be astronomically large, return it modulo 10⁹ + 7.

Input & Output

example_1.py — Basic Case

$ Input: n = 1, x = 2, y = 3

› Output: 6

💡 Note: With 1 performer and 2 stages: performer can go to stage 1 (3 possible scores) or stage 2 (3 possible scores). Total: 3 + 3 = 6 ways.

example_2.py — Multiple Performers

$ Input: n = 5, x = 2, y = 1

› Output: 32

💡 Note: Each performer can choose stage 1 or 2 (2^5 = 32 assignments). Since y = 1, each occupied stage gets score 1, so total is 32 ways.

example_3.py — Single Stage

$ Input: n = 3, x = 1, y = 4

› Output: 4

💡 Note: All performers must go to the single stage (1 way), and that stage can have 4 different scores. Total: 1 × 4 = 4 ways.

Constraints

1 ≤ n ≤ 1000 (number of performers)
1 ≤ x ≤ 1000 (number of stages)
1 ≤ y ≤ 10⁹ (maximum score value)
Answer must be returned modulo 10⁹ + 7

Visualization

Tap to expand

Understanding the Visualization

Performers Arrive

n performers need stage assignments

Stage Assignment

Each performer assigned to one of x stages

Band Formation

Performers on same stage form bands

Scoring Phase

Each occupied stage gets score 1 to y

Key Takeaway

🎯 Key Insight: Use inclusion-exclusion principle with Stirling numbers to count ways to partition performers into non-empty stage groups, then multiply by stage selection and scoring combinations

Asked in

G Google 45 f Meta 35 a Amazon 28 ⊞ Microsoft 22

This problem requires combinatorial mathematics using the inclusion-exclusion principle. The optimal approach computes Stirling numbers of the second kind to count ways to partition performers into non-empty groups, then multiplies by stage selection combinations and scoring possibilities. Time complexity: O(min(n,x)²)

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming with Inclusion-Exclusion	O(min(n,x)^2)	O(min(n,x)^2)	Use inclusion-exclusion principle to count valid arrangements mathematically

Dynamic Programming with Inclusion-Exclusion — Algorithm Steps

Calculate Stirling numbers of the second kind S(n,k) for distributing performers
Use inclusion-exclusion: ways with exactly k occupied stages = C(x,k) * S(n,k) * k!
For each k from 1 to min(n,x), multiply by y^k scoring possibilities
Sum all possibilities and return result modulo 10^9+7

Visualization

Tap to expand

Step-by-Step Walkthrough

Calculate Stirling Numbers

S(n,k) = ways to partition n performers into k non-empty groups

Choose Stages

C(x,k) ways to choose k stages from x available stages

Assign Groups to Stages

k! ways to assign k groups to k chosen stages

Score Occupied Stages

y^k ways to assign scores to k occupied stages

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

#define MOD 1000000007

long long modPow(long long base, long long exp, long long mod) {
    long long result = 1;
    while (exp > 0) {
        if (exp % 2 == 1) result = (result * base) % mod;
        base = (base * base) % mod;
        exp /= 2;
    }
    return result;
}

long long modInverse(long long a, long long mod) {
    return modPow(a, mod - 2, mod);
}

int min(int a, int b) {
    return a < b ? a : b;
}

long long combination(int n, int k, long long* fact) {
    if (k > n || k < 0) return 0;
    if (k == 0 || k == n) return 1;
    long long num = fact[n];
    long long den = (fact[k] * fact[n-k]) % MOD;
    return (num * modInverse(den, MOD)) % MOD;
}

int numberOfWays(int n, int x, int y) {
    // Precompute factorials
    long long* fact = (long long*)malloc((x + 1) * sizeof(long long));
    fact[0] = 1;
    for (int i = 1; i <= x; i++) {
        fact[i] = (fact[i-1] * i) % MOD;
    }
    
    // Compute Stirling numbers of second kind
    long long** S = (long long**)malloc((n + 1) * sizeof(long long*));
    for (int i = 0; i <= n; i++) {
        S[i] = (long long*)calloc(x + 1, sizeof(long long));
    }
    S[0][0] = 1;
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= min(i, x); j++) {
            S[i][j] = ((long long)j * S[i-1][j] + S[i-1][j-1]) % MOD;
        }
    }
    
    long long result = 0;
    for (int k = 1; k <= min(n, x); k++) {
        long long ways = (combination(x, k, fact) * S[n][k]) % MOD;
        ways = (ways * fact[k]) % MOD;
        ways = (ways * modPow(y, k, MOD)) % MOD;
        result = (result + ways) % MOD;
    }
    
    // Free memory
    free(fact);
    for (int i = 0; i <= n; i++) {
        free(S[i]);
    }
    free(S);
    
    return result;
}

int main() {
    int n, x, y;
    scanf("%d %d %d", &n, &x, &y);
    printf("%d\n", numberOfWays(n, x, y));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(min(n,x)^2)

Computing Stirling numbers takes O(n*x) time, and we iterate through O(min(n,x)) values

✓ Linear Growth

Space Complexity

O(min(n,x)^2)

Need to store Stirling numbers in 2D array

⚡ Linearithmic Space

27.4K Views

Medium Frequency

~35 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming with Inclusion-Exclusion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler