Question

i have 3 last problem.

1. in how many ways can 6 people be lined up to get on a bus if 3 specific persons insists on following each other?

2. in how many ways can 5 starting positions on a basketball team be filled with 8men who can play any of the positons?

3. how many 3 digit numbers can be formed from the digit 0,1,2,3,4,5, and 6, if each digit can be used only once?

Answer 1

1. If you consider the individual people as a group to themselves, and the three that follow together as a group, you have a total of four groups of people boarding the bus. You can arrange 4 groups in 4! ways. However, the three people that board together may not always go in the same order, so they can be arranged in 3! ways. In total then, you have 4!*3! ways to order the 6 people....

2.With 8 players and 5 spots, and you don't care about the order, this is a basic combination problem. There are 8C5 ways to choose the lineup.
The solution is 8!/(5!x3!)=56 ways......

3.the first digit can only be 1-6(6 possibilities), the second digit can be 0 or any of the remaing 5 numbers of 1-6 other than the one that is taken for first digit(6 possibilities), the third digit can be the remaining 5 digit from 0-6 other than the two taken for the first and second digit (5 possibilities)
6x6x5=180 numbers can be formed...

Answer 2

hope it helps @iramnaj