Question

In how many ways may one seat 100 people into 20 distinct round tables in such a
way that there are 5 people per table?​

Answer 1

Answer:

we can find that using the law of permutation. permutation can help us to find all the possible ways they can be seated

in the above question,

number of seats, r=20×5=100

number of people, n=100

so,

the ways they can be seated,

npr=100p100

Answer 2

Answer:

$\frac{100! * 20!*(4!)^{20} }{(5!)^{20}}$

Step-by-step explanation:

There are 20 Table on which 5 persons each are to be seated.

First Table(Select 5 from 100 and arrange) = [100C5] x (4!)

Second Table(Select 5 from Remaining 95 and arrange) = [95C5] x (4!)

Third Table(Select 5 from Remaining 90 and arrange) = [90C5] x (4!)

so on up till Last Table ....

20th Table(Select 5 from Remaining 5 and arrange) = [5C5] x (4!)

Now Cases when all are seated on table and arrange =

Multiplication of above = [100C5](4!) x [95C5](4!) x [90C5](4!)  x .. x [5C5](4!)

Now These 20 Tables are to be arranged among themselves =

Therefore Total Cases = {[100C5](4!) x [95C5](4!) x [90C5](4!)  x .. x [5C5](4!)} x (20!)

Which on solving equals to

$\frac{100! * 20!*(4!)^{20} }{(5!)^{20}}$