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

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SUMMARY

The discussion focuses on calculating the number of ways to form a security code using three letters from the alphabet set {a, b, c, d, e} and four digits from the set {1, 2, 3, 4, 5, 6}, with specific restrictions. The total number of combinations without restrictions is 756,000, while the number of combinations that include at least two consonants is 529,200. Participants clarify the use of permutations and combinations in their calculations, emphasizing the need to consider arrangements of the selected characters.

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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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