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

In summary: So the final answer is 5C3 x 6C4 x 7P7 = 756000In summary, the possible number of different ways a security code can be formed, using three letters and four numbers from the alphabet {a,b,c,d,e} and digits {1,2,3,4,5,6}, is 756000. This is calculated by first choosing 3 distinct letters and 4 distinct numbers, and then finding the number of ways to arrange those letters and numbers. This does not take into account any restrictions. Additionally, if the security code must contain at least two consonants, the number of possible combinations is 529200.
  • #1
iwan89
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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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  • #2
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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  • #3
why i can't use 5P3 x 6P4? you mean i should use C?
 
  • #4
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.
 
  • #5
You mean 5C3 x6C4?
 
  • #6
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?
 
  • #7
supposely letter has 5P3 and numbers has 6P4 but as a whole i don't know how to compute it
 
  • #8
How many ways can you rearrange 7 objects?
 
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  • #9
Got it thanks!
 

What is permutation with restriction?

Permutation with restriction is a mathematical concept that involves arranging a set of objects or elements in a specific order while also adhering to certain conditions or restrictions.

What are some common examples of permutation with restriction?

Some common examples of permutation with restriction include arranging students in a line according to their heights, organizing books on a shelf based on their genres, and creating a seating chart for a wedding based on familial relationships.

What is the difference between permutation with restriction and combination with restriction?

The main difference between permutation with restriction and combination with restriction is that permutation involves arranging objects in a specific order, while combination does not consider the order of the objects.

What are the key principles to keep in mind when solving permutation problems with restriction?

The key principles to keep in mind when solving permutation problems with restriction are identifying the number of objects, determining the number of ways the objects can be arranged, and taking into consideration any restrictions or conditions that need to be followed.

How can permutation with restriction be applied in real-life scenarios?

Permutation with restriction can be applied in real-life scenarios such as creating schedules for a sports team, arranging seating in a theater based on ticket prices, and organizing a menu based on dietary restrictions of guests.

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