Answer :
ANSWER :
The answers are :
5,773,185
is not
combination
EXPLANATION :
There are 110 seats.
and there are 110 ways to choose the first empty seat
Now, there are 109 seats left.
and there are 109 ways to choose the second empty seat
Now, there are 108 seats left
and there are 108 ways to choose the third empty seat
Now, there are 107 seats left
and there are 107 ways to choose the fourth empty seat
The total number of ways is the product of the result above :
110 x 109 x 108 x 107 = 138,556,440
We will divide this to 4! or 4 x 3 x 2 x 1 = 24
Since the order is not important.
That will be :
[tex]138,556,440\div24=5,773,185[/tex]Since the order is not important, this is a combination problem.
Final answer:
The question pertains to the subject of Mathematics for a High School grade level. It involves calculating combinations to determine the number of ways 4 seats can be left empty in a 110-seat auditorium. The calculation uses the combinations formula C(110, 4) to find the solution.
Explanation:
The subject of the question involves calculating the number of ways that 4 seats can be left empty in an auditorium that seats 110 people. This is a mathematical problem related to combinatorics, specifically involving combinations. Since the order in which the seats are left empty does not matter, we are dealing with combinations, not permutations.
To calculate the number of combinations, we use the formula:
C(n, k) = n! / [(n-k)! k!], where n is the total number of items (seats), and k is the number of items to choose (empty seats).
For our problem:
- n = 110 (total seats)
- k = 4 (empty seats)
The combinations formula will give us the number of ways to choose 4 empty seats from 110.
C(110, 4) = 110! / [(110-4)! 4!] = 110! / (106! 4!) = 4,598,126, 365 possible ways.
