Doing some how many combinations calculation

AI Thread Summary
The discussion focuses on calculating the number of unique combinations of the digits in the number 123215. The correct approach involves using permutations while accounting for the repeated digits, specifically the two 1s and two 2s. The calculation starts with the total permutations of six digits, which is 6!, and then divides by the factorial of the counts of the repeated digits (2! for the 1s and 2! for the 2s). This results in the final answer of 180 unique combinations. The conversation highlights the importance of understanding permutations in the context of repeated elements.
Compaq
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Doing some "how many combinations" calculation

Homework Statement



We're given the number 123215.

How many different numbers can we get by changing the order of the digits?

This is a six-digit number, so there are six positions for each digit.


This is an unsorted selection with covering (not sure if that's what you call it in English).



The Attempt at a Solution



I've tried nCr, nPr and pretty much everything that's tryable;) I just can't get it right. I know for a fact that the correct answer is 180.
 
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Do you know what nCr physically means? Well, it tells you how many ways you can select (without ordering) r different objects, from a set of n objects.

This is not what you have to do here. you have Order the given 6 digits, in order to produce newer numbers.

Order, and selection is done by nPr. Now you have to select and order 6 digits, from a set of 6 digits. So you must use 6P6.

But wait! That only works if all the digits are different. Here there are two 2's and two 1's. Can you account for that?
 


So I'm going to order a set of 6 digits. There are 4 different digits, but two 2s and two 1s. 6 digits total. The order of the 2s and the 1s doesn't matter.

My first guess was 6P4, but that can't be right.
 


Compaq said:
So I'm going to order a set of 6 digits. There are 4 different digits, but two 2s and two 1s. 6 digits total. The order of the 2s and the 1s doesn't matter.

My first guess was 6P4, but that can't be right.

Think it out logically. Let's look at the 1s only. Every number you create will contain these 2 digits whose order you can transpose without affecting the whole number. That means there are exactly two of these meaningless rogue permutations per whole number, so we can count properly by first counting all the permutations pretending each number is distinct and then dividing by 2. Can you see how to get rid of the rest of the rogues pertaining to the two 2s?
 


ahh, so we divide by 2 for the 1s, and then divide by 2 again for the 2s. That gives 180:D

Not really logic to me:S Probability is a weird branch of maths.

And thanks for the help! Appriciated it.
 
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