Question

Four republicans, 8 democrats, and 5 independents are randomly rated in a row. What is the probability that

1. the first 5 seats are taken by the democrats?

2. none of the first 5 seats is taken by democrats?

3. the final 3 seats are taken by a republican, and an independent

4.all republicans are seated together

Answer 1

4 Republicans, 8 democrats and 5 independents
total 17 people needs to be seated
This can be done in 17! ways

#1.
5 democrats for first 5 seats can be selected in 8C5 ways and they can be arranged in 5! ways among themseleves
Remaining 12 people can be arrange in 12! ways

Requried probability = (8C5 * 5! * 12!)/17!

#2.
First 5 seats can be selected in 9C5 ways and arrangements among themseleves can be done in 5! ways.

Remaining 12 people can be arranged in 12! ways

Requried probability = (9C5 * 5! * 12!)/17!

#3.
3 people for last 3 seats can be selected in 9C3 ways and they can be arranged in 3! ways among themselves

Remainig 14 people can be arranged in 14! ways

Requried probability = (9C3 * 3! * 14!)/17!

#4.
If all republicans are seated together, there will be 1 + 8 + 5 = 14
These can be arranged in 14! ways and 4 republicans can be arranged in 4! ways among themselves

Required probability = (14! * 4!)/17!