Ordering of 'N' Distinguishable Objects

[Hart]

Homework Statement

Suppose I have 'N' distinguishable objects.

1. In how many different ordered sequences can they be arranged? And why?

2. In how many ways can they be split up into two piles?

(ordering within the piles being unimportant)

The first pile to contain 'n' objects and the second 'm', with therefore (n + m = N).

Homework Equations

Stated within the solution attempt.

The Attempt at a Solution

1. Obviously the answer is $$N!$$ , but I need to show this and not just state it. I'm having trouble explaining how to prove the result. I went along the lines of:

• If I have 2 objects (n=2) then there are 2 combinations / possible arrangements (1&2 or 2&1).
  
• Alternatively, there are 2 possibilities for the first term in the sequence and then only 1 possibility for the second / final term in the sequence.
  
• This can be represented as (2)(1) = 2 = 2!

• If I have 3 objects (n=3) then there are 6 combinations / possible arrangements (1&2&3 or 1&3&2 or 2&1&3 or 2&3&1 or 3&1&2 or 3&2&1).
  
• Alternatively, there are 3 possibilities for the first term in the sequence, then 2 possibilities in the second term, then only 1 possibility in the final term.
  
• This can be represented as (3)(2)(1) = 6 = 3!

Hence generalising to N objects:

• If I have N objects (n=N) then there are multiple combinations / possible arrangements.
  
• Alternatively, there are N possibilities for the first term in the sequence, then N-1 possibilities in the second term, then N-2 possibilities for the next term, then continually down until only 1 possibility in the final term, i.e.
  
  $$(N)(N-1)(N-2)...(N-i)...(2)(1)$$
  
  where i is a positive integer value.
  
• So: $$(N)(N-1)(N-2)...(N-i)...(2)(1) = N!$$
  
  
  Which gives the result, but I think the proof could be much better.. I know it's not that difficult, but I'm struggling with how to show it (i.e. the notations, equations, etc) and could do with some advice/help/direction on the best way to write this all down?
  
  2. I was advised that this follows on from knowing part 1, and that the answer was not (N-1). But I don't have an idea how to calculate this at the moment.

 

[kuruman]

Imagine having a bag with N balls, each a different color from all others. You stick your hand in the bag and pull out a ball.

Q1: How many different ways are there that you can do this?
A1: N

Now there are N-1 balls left in the bag. Just like before, if you stick your hand in the bag again there are N-1 ways you can pull out a second ball.

Q2: How many ways total are there that you can pull two balls out of the bag if you care about the order in which they are pulled out?
A2: N(N-1) because for every one ball that you pulled out first, there are N-1 balls that you can pull out next.

Do you see how it goes?