What is the probability that a random rearrangement of the letters in the word "mathematics" will begin with the latters "math"?
Probability = (# of desired results) / (# of total results)
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?