Math Is Fun Forum

nsl22

Member

Registered: 2007-08-16

Posts: 8

License Plate type Permutation problem

Here is the problem that I cannot figure out .  Please help me if you can. 

Ten different letters of an alphabet are given.  2 of these letters followed by 2 digits are used to number the products of a company.  In how many ways can the products be numbered? 

The answer is 1000. 

I'm confused by it because in order for it to be a valid combination, it has to be in the form of LL## where L = letter and # = number.  If you say that for the first letter you have 10 choices, 2nd letter you ahve 10 choices, 1st number 10 choices and 2nd number 10 choices, you come up with 10*10*10*10 = 10,000 which is wrong because this is including the combinations of #L#L, ##LL, and L#L#.   

How do you figure this out for only LL## combinations? 

Thanks a lot!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: License Plate type Permutation problem

The answer is indeed 10000. There are 10000 distinct LL## permutations. If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher – 60000 to be precise.

Last edited by JaneFairfax (2007-08-16 09:17:46)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: License Plate type Permutation problem

If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher – 60000 to be precise.

Wouldn't it be 160,000?  20*20*20*20?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

nsl22

Member

Registered: 2007-08-16

Posts: 8

Re: License Plate type Permutation problem

The answer is indeed 10000. There are 10000 distinct LL## permutations. If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher – 60000 to be precise.

Can you explain?  To me, it doesn't seem like you can have more than 10,000 combinations without getting duplicates.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: License Plate type Permutation problem

If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher – 60000 to be precise.

Wouldn't it be 160,000?  20*20*20*20?

You are only allowed two letters and two numbers.

The answer is indeed 10000. There are 10000 distinct LL## permutations. If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher – 60000 to be precise.

Can you explain?  To me, it doesn't seem like you can have more than 10,000 combinations without getting duplicates.

Why not?

Try it with just two letters (A, B) and two numbers (0, 1). That might give you the general idea. With only two letters and two numbers, there are 2 × 2 × 2 × 2 = 16 permutations in the order LL##. If you allow #L#L, ##LL, L#L#, etc as well, the number of permutations is 6 × 16 = 96.

Last edited by JaneFairfax (2007-08-16 11:11:28)

nsl22

Member

Registered: 2007-08-16

Posts: 8

Re: License Plate type Permutation problem

That makes sense to me now.  I read your simplified explanation and understood everything but where the 6 came from. 

I figured it out though 6 = 4!/(2!*2!)
since there are 4 spaces and 2 #'s and 2 L's.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: License Plate type Permutation problem

You’ve got it. The 6 is from [sup]4[/sup]C[sub]2[/sub].

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: License Plate type Permutation problem

Jane - I came up with 60000 also if you allow orders other than LLNN but I was wondering if there was an easier way to calculate, i.e. a simple formula.   I calculated the number of ways where neither number or letter were repeated such as AB12 (48600), if either the letter or the number was repeated such as AA12 or AB11 (5400 each), and if both the letter and the number was repeated such as AA11 (600).

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: License Plate type Permutation problem

You can make a four variable hypercube with length nine on each side.
Then there are ten points, one unit apart in all 4 directions.
The ten cubes are inside each other getting smaller, if you
represent 4-D in 3-D.
Your 4 directions are:
  1.  in-out
  2.  left-right
  3.  up-down
  4.  back-front

Last edited by John E. Franklin (2007-08-20 11:27:22)

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