Question 1194573: In how many ways can 4 people be seated in a row of 12 chairs
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's say the four people have the code names of A,B,C,D
We'll have twelve slips of paper with the numbers 1,2,3,...,10,11,12 written on them (one number per paper).
Each number refers to a different chair.
The process is that each person will pick a slip of paper at random out of a hat.
Whatever is chosen is not put back or replaced. In other words, repeats are not allowed.
One possible permutation is: 12,8,7,1
This means:
person A is in chair 12
person B is in chair 8
person C is in chair 7
person D is in chair 1
As you can see, we have n = 12 slips of paper to pick from and r = 4 slots to fill. The order matters.
That means you'll compute nPr = 12P4 as shown below.
n P r = (n!)/( (n-r)! )
12 P 4 = (12!)/( (12-4)! )
12 P 4 = (12!)/( 8! )
12 P 4 = (12*11*10*9*8!)/( 8! )
12 P 4 = 12*11*10*9
12 P 4 = 11880
or slightly alternatively
n P r = (n!)/( (n-r)! )
12 P 4 = (12!)/( (12-4)! )
12 P 4 = (12!)/( 8! )
12 P 4 = (12*11*10*9*8*7*6*5*4*3*2*1)/( 8*7*6*5*4*3*2*1 )
12 P 4 = (479,001,600)/( 40,320 )
12 P 4 = 11880
Answer: 11880
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
First person can seat at any of 12 seats, giving 12 options.
Second person can seat at any of remaining 11 seats, giving 11 options.
Third person can seat at any of remaining 10 seats, giving 10 options.
Fourth person can seat at any of remaining 9 seats, giving 9 options.
The total number of possible seating arrangements is 12*11*10*9 = 11880. ANSWER
It is the product of 4 consecutive integers numbers in descending order, starting from the number of 12.
Solved and explained.
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To see many similar (and different) solved problems, look into the lesson
- Special type permutations problems
in this site. Learn the subject from there.
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