porcupinetree
not actually a porcupine
- Joined
- Dec 12, 2014
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- HSC
- 2015
Re: 2015 permutation X2 marathon
Case 1 - 2 gates have 2 cars while the other 6 gates have 1 car
8C2 - choose 2 gates
10C2*2! - choose 2 cars for gate 1 & arrange
8C2*2! - choose 2 cars for gate 2 & arrange
6! - arrange other cars
Total for case 1 = 8C2.10C2.2!.8C2.2!.6! = 101606400
Case 2 - 1 gate has 3 cars, the rest have 1
8C1 - choose the gate to have 3 cars
10C3.3! - choose 3 cars for gate 1 and arrange
7! - arrange the other cars
Total for case 2: 29030400
Total = 101606400 + 29030400 = 1 306 636 800
Here's what I have (assuming the order of cars through the gates matters):There are 8 gates which 10 cars can pass through. How many ways can the cars pass through if every gate is used?
(Sorry my initial answer was incorrect. Working out another one.)
Case 1 - 2 gates have 2 cars while the other 6 gates have 1 car
8C2 - choose 2 gates
10C2*2! - choose 2 cars for gate 1 & arrange
8C2*2! - choose 2 cars for gate 2 & arrange
6! - arrange other cars
Total for case 1 = 8C2.10C2.2!.8C2.2!.6! = 101606400
Case 2 - 1 gate has 3 cars, the rest have 1
8C1 - choose the gate to have 3 cars
10C3.3! - choose 3 cars for gate 1 and arrange
7! - arrange the other cars
Total for case 2: 29030400
Total = 101606400 + 29030400 = 1 306 636 800