60 wedding guests are to be seated at 10 circular tables, each with 6 seats. Two seating arrangements are considered the same if the person to the left of each person is unchanged. How many seating arrangements are possible, and why? It is ok to leave your answer with factorials in it.

Step-by-step explanation:

60 guests are there with 10 circular tables of 6 seats each.

First we have to select 6 persons from 60 for I table and next 6 persons for II table etc.

i.e.first table can be assigned 6 people in 60C6 ways

II table from remaining 54 in 54C6 ways

....

Each table with 6 persons can be arranged in a circular table in 5! ways

(since Two seating arrangements are considered the same if the person to the left of each person is unchanged)

Thus no of ways of allotting persons to different tables

= 60C6*54C6*48C6*42C6...*6C6

Each table allotment is in (6-1)! = 5! ways

Total ways of seating arrangement =

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

i hope this is what you were looking for.

see attached graph