One of the class objectives is to give an oral presentation to the professor. This time it has to do with explaining Permutations and Combinations. We have 4 things we need to explain:(adsbygoogle = window.adsbygoogle || []).push({});

1) Permutations / Repetitions are not allowed / Order Matters

2) Combinations / Repetitions are not allowed / Order Doesn't matter

3) Permutations / Repetitions are allowed / Order Matters

4) Combinations / Repetitions are allowed / Order Doesn't matter

Here is what I have so far as far as explainations go:

1) P(n,k), n = number of objects, k = pickings

So we permute n, the number of objects... n!, this gives us n(n-1)(n-2)...(n-(k-1))*(n-k)!, so we have the total permutations of n objects. Now we must divide by (n-k)!, since these elements are ones past the kth element.

This gives us the formula n! / (n-k)!

2) C(n,k) n choose k.

We are permuting the n! / (n-k)!, this will give us the permutation of n objects on k elements. Since we cannot have combinations such as 256 and 562, we must eliminate them. We do so by multiplying k! with (n-k)!; so we get n! / (n-k)!k!. My problem here is I can't understand why that k! is there... I know it has something to do how the thing gets ordered, but I can't explain it. Maybe somebody can help me out here.

3) Example: AABBCA;

The formula we get is... (m1 + m2 + ... mn)! / (m1!*m2!*...*mn!); and I need to explain how we get that...

Here we have 3 objects, n = 3.

We permute all the 'copies' of each object. (3 A's + 2 B's + 1 C)!. This gives us the total permutations. (I need some help explaining that a bit better). What does this 'give us'?...

Then I know we take that and then we permute each object, by the number of copies, so since we have 3 A's we do 3!, for the B's we have 2!, and for C we have 1!. I by the multilplication principle we multiply them together, 3!*2!*1!; so then we (3 A's + 2 B's + 1 C)! / 3!*2!*1!. I'm a bit lost as to why we divide... I know that it's because order matters, so we can have 245 and 254, they are not the same, and repitions are allowed... But why?...

4) Here we'll use the 00|000|0, where

k = number of circles

number of "|" = n-1

How many 'types' of objects do we have?... no f'en clue.

But in the end we plug in k for m1 (in the previous equation), and plug in n-1 for m2 from the previous equation (no clue as to why...) and we get...

(k+(n-1))! / (k!(n-1)!)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: [Discrete Math] Permutations / Combinations Advice needed

**Physics Forums | Science Articles, Homework Help, Discussion**