Combinatorics simple problem
Maths and statistics discussion, revision, exam and homework help.
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Combinatorics simple problem1. You’ve just established an account with eatmybroccoli.com and you need a password. The password
must be exactly 6 characters long. Each character must be a lower-case letter (a through z) or a
number (0 through 9). Exactly three characters must be numbers and exactly three characters must
be letters, but they can be in any order. No character can be repeated in the password. How many
different passwords can you make?
My answer would be:
That's how I see it: we can choose 3 letter out of 26 (a-z) in
ways and 3 digits out of 10 (0-9) in 
ways. Now, any of the 6 characters can take any of the 6 places available, thus 
. All combined, gives the answer given above.
Please tell me, what do I do wrong here?
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Re: Combinatorics simple problemYou have an alphabet of 36 characters to choose from: {a, .. z, 0 .. 9} so your calculation will start(Original post by joanna1996)
1. You’ve just established an account with eatmybroccoli.com and you need a password. The password
must be exactly 6 characters long. Each character must be a lower-case letter (a through z) or a
number (0 through 9). Exactly three characters must be numbers and exactly three characters must
be letters, but they can be in any order. No character can be repeated in the password. How many
different passwords can you make?
My answer would be:
That's how I see it: we can choose 3 letter out of 26 (a-z) in ways and 3 digits out of 10 (0-9) in ways. Now, any of the 6 characters can take any of the 6 places available, thus . All combined, gives the answer given above.
Please tell me, what do I do wrong here?
etc
The letters aren't being treated separately to the digits. Does that help?
[oops: hold the press: I misread the question - it's restricted to 3 numbers and 3 letters, in which case, your approach looks fine at first sight]Last edited by atsruser; 03-08-2012 at 13:11. -
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Re: Combinatorics simple problemYou're overcounting the number of permutations. You can take 26C3 different groups of 3 letters and 10C3 different groups of 3 numbers to give 26C3 x 10C3 x 6! = 224640000 perms.(Original post by joanna1996)
1. You’ve just established an account with eatmybroccoli.com and you need a password. The password
must be exactly 6 characters long. Each character must be a lower-case letter (a through z) or a
number (0 through 9). Exactly three characters must be numbers and exactly three characters must
be letters, but they can be in any order. No character can be repeated in the password. How many
different passwords can you make?
My answer would be:
That's how I see it: we can choose 3 letter out of 26 (a-z) in ways and 3 digits out of 10 (0-9) in ways. Now, any of the 6 characters can take any of the 6 places available, thus . All combined, gives the answer given above.
Please tell me, what do I do wrong here?
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Re: Combinatorics simple problemFirst,(Original post by joanna1996)
So which answer would be correct? The official one says it should be:
, but I don't understand why 
should be there...
and 
so 
counts the total number of ways we can combine all perms of 3 letters with all perms of 3 numbers. For example, we can do the following:
{a,b,c} with {1,2,3}
{a,b,c} with {1,3,2}
...
{b,a,c} with {1,2,3}
{b,a,c} with {1,3,2}
...
[many, many combinations of sets omitted]
...
{x,y,z} with {1,2,3}
{x,y,z} with {1,3,2}
..
{z,y,x} with {3,2,1}
The left hand column holds all the possible perms of 3 from 26 letters, the right hand column holds all the possible perms of 3 from 10 numbers.
If we take a particular combination, say {p,a,e} with {6,1,4}, we are not allowed to change the order of the letters or numbers (as we've already made a list of all possible orders) but we can order them left-to-right in six boxes, like so:
p,a,e,6,1,4
p,a,6,e,1,4
p,6,a,1,4,e
and so on. How many ways can we do this for each pair of permutation sets?
Well, we have 6 boxes, and once we have chosen 3 of them for the letters, we have by default chosen the remaining 3 for the numbers. Thus, we can make
different arrangements using any given pair of each of the permutation sets. Hence we have a total of arrangments.
Does that make sense? Personally, I prefer the argument I gave earlier, but it's useful to do it in different ways. -
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Re: Combinatorics simple problemMy pleasure.(Original post by joanna1996)
Yes, now it makes a lot of sense!
Thank you very much! :*
I must say I also prefer your version -
is just a lot clearer to me than the official suggestion. I'm very grateful for your help
I ought to point out that my notation was very loose in the explanation above. I was holding different permutations in things that I referred to as "sets"; this is mathematical nonsense, since sets don't remember the order of the objects inside them. I should really have called them ordered lists, or something. That doesn't change the argument though. -
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Re: Combinatorics simple problemI'm sorry, I'm not from the UK, so I'm not doing A-levels and have unfortunately no idea to which model do the problems of this type belong.(Original post by SubAtomic)
What module is this please?
Just studied something similar but I couldn't have answered your question. I obviously need more understanding.
I can only say that I found this question (#26) here: http://mcis.jsu.edu/mathcontest/comp1ans12.pdf -
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Re: Combinatorics simple problemIt's ok thanks, it is part of Binomial theorem, so it could be in C2 or C4, but maybe it is to do with synthesising all that is learnt, mind you America will have different courses so that would help when trying to answer a question from an Alabama maths competition(Original post by joanna1996)
I'm sorry, I'm not from the UK, so I'm not doing A-levels and have unfortunately no idea to which model do the problems of this type belong.
I can only say that I found this question (#26) here: http://mcis.jsu.edu/mathcontest/comp1ans12.pdf
.
Thank you very much! :* 
