Outside a home, there is a 5-key keypad with letters A, B, C, D and, E that can be used to open the garage if the correct five-letter code is entered. Each key may be used only once. How many codes are possible? The number of possible codes is C
Outside a home, there is a 5-key keypad with letters A, B, C, D and, E that can be used to open the garage if the correct five-letter code is entered. Each key may be used only once. How many codes are possible? The number of possible codes is C
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Please do not write the answer in step 1 or step 2 boxes. The website doesn't allow me to see the full explanatin. Could you write it on paper and then post it as a pic? Please.
![**Problem:**
Outside a home, there is a 5-key keypad with letters A, B, C, D, and E. This keypad can be used to open the garage if the correct five-letter code is entered. Each key may be used only once. How many codes are possible?
---
**Solution:**
To find the number of possible five-letter codes, you must consider the permutations of the five distinct letters: A, B, C, D, and E.
Since each key can only be used once, the number of arrangements (permutations) of the letters is determined by calculating the factorial of the number of letters.
There are 5 letters, thus the number of different codes is:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Therefore, the number of possible codes is **120**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38aa5034-1eec-4964-a430-19a44c4ef4d5%2F6c387a06-0099-45be-90d9-b179f79fc3b6%2Fxdjio9o_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Outside a home, there is a 5-key keypad with letters A, B, C, D, and E. This keypad can be used to open the garage if the correct five-letter code is entered. Each key may be used only once. How many codes are possible?
---
**Solution:**
To find the number of possible five-letter codes, you must consider the permutations of the five distinct letters: A, B, C, D, and E.
Since each key can only be used once, the number of arrangements (permutations) of the letters is determined by calculating the factorial of the number of letters.
There are 5 letters, thus the number of different codes is:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Therefore, the number of possible codes is **120**.
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