I miss when everyone always said this
good times but ty for the points
By increasing the number of blue widgets supplied
Answer:
There are 1% probability that the last person gets to sit in their assigned seat
Step-by-step explanation:
The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.
If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.
LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!
Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.
P = 99!/100! = 99!/ (100 * 99!) = 1/100
There are 1% probability that the last person gets to sit in their assigned seat
8X + 8 > 7x + 14
subtract 7x and 8 from both sides
x > 6
Answer: 2405970 ways
Step-by-step explanation:
When the order of selecting r individuals out of n individual , we use permutations.
The number of permutations of r things taking out of n things is given by :-
Given : There are 135 people and three door prizes .
i.e. n= 135 , r= 3
Then, the total number of ways = 
Hence, there are 2405970 ways to distribute door prizes of $7,500, $750 and $75 .
