Doing some how many combinations calculation

In summary, there are six positions for each digit in the number 123215, and that number can be produced by Ordering the given 6 digits, in order to produce newer numbers. The number 180 can be produced by dividing the number 123215 by 6 and then counting the number of different ways that it can be done.
  • #1
Compaq
23
0
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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  • #2


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?
 
  • #3


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.
 
  • #4


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?
 
  • #5


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.
 

FAQ: Doing some how many combinations calculation

How do you calculate the number of combinations?

The number of combinations can be calculated using the formula nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen.

What is the difference between combinations and permutations?

Combinations refer to the number of ways a group of items can be selected without considering the order, while permutations take into account the order in which the items are selected.

Can combinations be used in real-life situations?

Yes, combinations are commonly used in real-life situations such as in probability, genetics, and combination locks. They can also be used to calculate the number of possible outcomes in a given scenario.

Are there any limitations to using the combination formula?

The combination formula assumes that all items are unique and that the order of selection does not matter. It may not be applicable in situations where these assumptions do not hold.

How can I check my combination calculation?

You can use a combination calculator or manually calculate the combinations to double-check your answer. You can also use the combination formula to solve different combinations problems to practice and improve your calculation skills.

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