# Math counting problem

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To open a safe, 4 number buttons must be pressed, in the correct order. Over time, the 4 numbers buttons of the code fade. A thief notices the faded buttons, so knows that the code consists of those 4 numbers.

How many possible codes are there?

4 numbers can be arranged in 4! different ways, so there are 24 different possible combinations.

The safe owner decides that he wants to change the code. How many possible codes are possible if each code can contain only 1 of the 4 faded numbers?

$$^4 C_1 \times ^8 C_3 \times 4!=5376$$

Since there are two groups, faded and non faded. One is selected from the faded numbers, and 3 from the remaining 8 non faded. i then multiplied by 4! since there are 4! ways of arranging the 4 numbers.

The answer guide says this is wrong though. Is anybody able to explain where i went wrong in my reasoning?

Thanks,
Dan.

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Kurdt
Staff Emeritus
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At first glance I would say that it has to do with the fact that you think 8 are non-faded. If there is a button for each digit then there will 10 buttons, 4 of which have faded. Thus there will be 6 non-faded buttons.

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The buttons arent 0-9, theyre 1-12. There was a diagram with the question that showed this.

Dick