# Python Program for Dining Table Seating Arrangement Problem

## Dining Table Seating Arrangement Problem

One of the real-world problems asked in TCS CodeVita where we need to find the impactful seating arrangements for the N number of attendees. Here is the Python Program for Dining Table Seating Arrangement Problem, which takes user input R for no of Round tables and N for the total number of attendees.

### Problem Statement

In a Conference ,attendees are invited for a dinner after the conference.The Co-ordinator,Sagar arranged around round tables for dinner and want to have an impactful seating experience for the attendees.Before finalizing the seating arrangement,he wants to analyze all the possible arrangements.These are R round tables and N attendees.In case where N is an exact multiple of R,the number of attendees must be exactly N//R,,If N is not an exact multiple of R, then the distribution of attendees must be as equal as possible.Please refer to the example section before for better understanding.
For example, R = 2 and N = 3
All possible seating arrangements are
(1,2) & (3)
(1,3) & (2)
(2,3) & (1)
Attendees are numbered from 1 to N.

Input Format:

• The first line contains T denoting the number of test cases.
• Each test case contains two space separated integers R and N, Where R denotes the number of round tables and N denotes the number of attendees.

Output Format:

Single Integer S denoting the number of possible unique arrangements.

Constraints:

• 0 <= R <= 10(Integer)
• 0 < N <= 20 (Integer)
Sample Input 1:
1
2 5
Sample Output 1:

10

Explanation:

R = 2, N = 5

(1,2,3) & (4,5)

(1,2,4) & (3,5)

(1,2,5) & (3,4)

(1,3,4) & (2,5)

(1,3,5) & (2,4)

(1,4,5) & (2,3)

(2,3,4) & (1,5)

(2,3,5) & (1,4)

(2,4,5) & (1,3)

(3,4,5) & (1,2)

Arrangements like

(1,2,3) & (4,5)

(2,1,3) & (4,5)

(2,3,1) & (4,5) etc.

But as it is a round table,all the above arrangements are same.

```t=int(input())
for T in range(t):
r,n=map(int,input().split())
ans=0
maxi = (n//r + (n%r!=0))
last= n//r
ans = maxi * (r-1) * last + 1
if (last!=1):
ans+=(maxi *(maxi-1))//last
print(ans)```

## C

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## C++

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## Java

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