# Rearrangement of the letters "mathematics" (probability)

1. Jun 29, 2016

### ecoo

1. The problem statement, all variables and given/known data

What is the probability that a random rearrangement of the letters in the word "mathematics" will begin with the latters "math"?

2. Relevant equations

Probability = (# of desired results) / (# of total results)

3. The attempt at a solution

The solution I got was (2*2*2*1) / (11*10*9*8), because there are 2 ways to choose "M" for the first letter, 2 ways to choose "A" for the first letter, and so on. This solution is supposedly correct, but I thought about it a bit more and I confused myself.

Here's my confusion:

If we think about all of the combinations of the word "mathematics", it would seem as if words that do not begin with the doubles (M, A, or T) have less combinations.

So for a combination that begins with "ICSE", one combination would be "ICSE"MATMATH. But would the combination "ICSE"MATMATH (looks the same, but interchange the M's, A's, and T's) also be considered unique?

And then you have combinations that begin with "MATH" (begins with the doubles), and an example is "MATH"EMATICS. As seen, the EMATICS portion has 7! unique combinations, which would certainly be more than the number of combinations in the first example above, if we choose not to distinguish between the two M's, A's, and T's.

And so I imagined all the combinations of the word "mathematics", and it would seem like there are less combinations that begin with non-doubles, if we don't distinguish between the doubles. So the chances of a combination of words beginning with "MATH" is slightly more 1/990, because there are more combinations that begin with "MATH", as opposed to combinations with 4 other letters.

I suppose it boils down to whether we consider the M's, A's, and T's to be unique in the problem. If they are unique, then the answer I got initially is correct. If they are not unique, then the answer would be slightly above 1/990. Is this conclusion correct?

Thanks

2. Jun 29, 2016

### ecoo

Thought about it a bit more... since I used 2*2*2*1 for my combinations of "MATH", then I already assumed that the M's, A's, and T's are unique, so it's all good.

3. Jun 29, 2016

### Delta²

You got me confused also with your thinking , however I surely think you do one mistake. The number of total results is not just (11*10*9*8), it is rather

$\frac{11!}{2!2!2!}$ if the 2 Ms, As , Ts are considered to be the same thing,

or $11!$ if you consider the each of the 2 Ms,As,Ts as unique.

4. Jun 29, 2016

### Delta²

Ok I think I see now, you have simplified the common factor 7! in numerator and denominator.

Seems correct IF we consider each of the similar letters as unique.

However something tells me that the exercise doesn't want us to consider those letters as unique.

5. Jun 29, 2016

### Ray Vickson

I don't know exactly what you mean by "unique". There are 11 letters altogether, which we can imagine are written out on 11 separate pieces of paper with a unique number from1 to 11 on one side of each piece and individual letters on the other side. Two of the pieces have "m" on one side, two have "a", etc. The pieces with "m" on them are physically separate but indistinguishable in terms of their outcome: picking either of them first will be equally good, and we don't care which of the two comes first. Still, they are separate objects.

Last edited: Jun 29, 2016
6. Jun 29, 2016

### Delta²

And the funny thing is that if we don't consider the M,A,T as unique, but separate objects that give the same outcome as Ray Vickson says, then the probability I get is

$\frac{7!}{\frac{11!}{2!2!2!}}$

which ends up to be the same number as with the unique case... Am I doing something wrong or is it just some coincidence (not a pure coincidence but because all of the double letters appear exactly once in the "constant" part of the word that is the "MATH" part)?

7. Jun 29, 2016

### Ray Vickson

I still have no idea what you mean by "unique": please explain. I know the meaning of the word "unique" in other contexts, but I cannot see how to apply it here.

8. Jun 29, 2016

### Delta²

Unique means that each M (or A or T) is considered to be different than the other M. So for example the word MATHEMATICS will appear (in all the possible rearrangements) 2 times (if we exchange the 2Ms) and another 2 times for the 2 As and another 2 times for the two Ts. Total 8 times (it ll be 2x2x2 and not 2+2+2 cause it goes multiplicative not additive). If we don't consider them unique these 8 times will count just as 1.

9. Jun 29, 2016

### Ray Vickson

Basically, you are describing the differences between so-called "classical statistics" and "Bose-Einstein" statistics (which applies to quantum particles with even spin numbers). In the weird world of atomic or subatomic systems such considerations become very important, but for objects at the macroscopic level (such as coins, dice, people, poker chips, circuit boards, etc.) the applicable "statistics" are classical. (This uses the word 'statistics' in an unusual way, in accordance with the way it is used in Quantum and/or Statistical Mechanics.) In classical statistics, identical objects are distinguishable in principle (even though may not bother to distinguish between them); for example, they may be physically different poker chips having the exact same monetary value. That is why you can count them separately. However, for Bose-Einstein objects, if you have three of them they cannot be distinguished in any way, so it makes no sense to speak of object 1, object 2 and object 3; you just have three objects.

Last edited: Jun 29, 2016
10. Jun 29, 2016

### ecoo

Hey guys, thanks for the help. The answer 1/990 is correct. The reason I got confused was because I focused on, not rearranging the word "mathematics", but the probability of having the first four letters as "MATH". When considering uniqueness as a factor, then the method works. But when assuming uniqueness is not a factor, the method I used does not work, because of the reasons I stated in the original post.

Just a note, for probability problems that pertain to, for example, rearrangement of letters, it often doesn't matter whether uniqueness is considered or not. As long as you stay consistent (e.g. if uniqueness is considered for the # of desired results, then uniqueness must also be considered for the # of total results), the answer will be correct. This is because if you do not consider certain arrangements as unique, then the factor that you divide the results by cancels out.

Last edited: Jun 29, 2016