Ways of Selection Problem

• Aug 8th 2010, 04:43 AM
Lukybear
Ways of Selection Problem
Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.

How many different codes are possible if the order is unimportant, and the digits can be repeated?

The answer here has 8 x 8 x 8.

But I think this is wrong, as this orders security code. I.e. its counting repeats.

For example, it would have counted 8 8 7 as different to 7 8 8 which will still work.

Am I flawed in my reasoning? If not how do i remove the ordering.
• Aug 8th 2010, 05:15 AM
Plato
Quote:

Originally Posted by Lukybear
Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.
How many different codes are possible if the order is unimportant, and the digits can be repeated?
The answer here has 8 x 8 x 8.
But I think this is wrong, as this orders security code. I.e. its counting repeats.

If the question actually states that order is unimportant then that answer is not correct.

The answer then is $\displaystyle \binom{3+8-1}{3}$.
That is simply selecting three from eight choices.
• Aug 8th 2010, 10:33 PM
Lukybear
But that dosent count repeition. I.e. the nos could be 888.
• Aug 9th 2010, 03:27 AM
Plato
It most certainly does count repetitions. It is the multi-selection formula.
• Aug 9th 2010, 05:23 AM
Lukybear
So sorry, Ive miss read your solution. I thought it was 8C3 (combinations) which was a bit strange.

But I do not understand how you acquired the 10C8 whic does not equate to 8C3.
• Aug 9th 2010, 07:00 AM
Plato
Go down the page to the section on Multiset coefficients.
• Aug 9th 2010, 07:56 AM
Quote:

Originally Posted by Lukybear
Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.

How many different codes are possible if the order is unimportant, and the digits can be repeated?

The answer here has 8 x 8 x 8.

But I think this is wrong, as this orders security code. I.e. its counting repeats.

For example, it would have counted 8 8 7 as different to 7 8 8 which will still work.

Am I flawed in my reasoning? If not how do i remove the ordering.

Suppose the pads are numbered 1 to 8.

Why is the solution not

$\binom{8}{3}+8(8)\ ?$

Ah!!

I get it!