How many ways can a security code be formed with restrictions?

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Homework Help Overview

The problem involves forming a security code using three letters from a set of alphabets {a,b,c,d,e} and four digits from {1,2,3,4,5,6}, with the condition that all characters can only be used once. The discussion includes two specific cases: one without restrictions and another requiring at least two consonants.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of permutations and combinations to determine the number of ways to select and arrange letters and digits. There is uncertainty about whether to use permutations or combinations and how to account for the arrangement of letters and digits together.

Discussion Status

The discussion is ongoing, with participants questioning the validity of their approaches and seeking clarification on how to properly compute the arrangements. Some guidance has been offered regarding the distinction between permutations and combinations, but no consensus has been reached on the correct method to apply.

Contextual Notes

Participants express confusion over the requirements for arrangement and selection, particularly in relation to the restrictions imposed in the second part of the problem. There is also mention of the need to consider how letters and digits can be combined in the final arrangement.

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Homework Statement



A security code is formed by using three alphabet and four digits chosen from alphabet {a,b,c,d,e} and digits {1,2,3,4,5,6}. All digits and alphabets can only be used once. Find the number of different ways the security code can be formed if
(a) there is no restriction imposed (Answer :756000)
(b) It consists of at least two consonant (Answer: 529200)


Homework Equations


I used Permutation with restriction

The Attempt at a Solution


i tried the following. I am not sure whether the working is correct or not..

(a) 5P3(for alphabet) x 6P4(for digit) =21600.. but this seems wrong because my computation is based on the alphabet and digits must be togather

(b) I have no idea at all. I really need the idea from the first question :(
 
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iwan89 said:
(a) 5P3(for alphabet) x 6P4(for digit) =21600.. but this seems wrong because my computation is based on the alphabet and digits must be togather
First, calculate how many ways there are to choose the letters and numbers. How many combinations are there if you choose three letters and four numbers without repetition?

Then, calculate how many ways there are to arrange each combination.
 
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why i can't use 5P3 x 6P4? you mean i should use C?
 
iwan89 said:
why i can't use 5P3 x 6P4?
Because that would give you the number of permutations of the form abc1234 or bec3462, where the letters come before the numbers. It would not include permutations like 12ab34c.
 
You mean 5C3 x6C4?
 
iwan89 said:
You mean 5C3 x6C4?
That gives you the number of ways to choose 3 distinct letters and 4 distinct numbers. Now given a particular choice, how many ways are there to rearrange those letters and numbers?
 
supposely letter has 5P3 and numbers has 6P4 but as a whole i don't know how to compute it
 
How many ways can you rearrange 7 objects?
 
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Got it thanks!
 

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