Question

How many different 3-letter strings can be formed from the letters of MATHEMATICS (no letter can be used in a given string more times than it appears in a word)?
a) 336
b) 399
c) 660
d) 675
e) 990

Answer 1

11!/2!2!2!7!=11!/8x7!=11!/8!=11x10x9=990

Answer 2

That's what I had too, but what about the repeated letters?

Answer 3

yes I've considered that already.

Answer 4

notice the 2!2!2! at the bottom

Answer 5

It doesn't seem like it should be the same as \[11^{P}3\] which also equals 990.
I found an answer of 399 on the internet.
\[8^{P}3\] for the permutations with no repeats + 3*3*7 for the permutations with repeats. Does that seem correct?

Answer 6

From: http://www.pc.maricopa.edu/data/GlobalFiles/file/mathematics/AMATYC%20Test/AMATYC%20SML%20Solutions%20Spring%202007.pdf
There are three pairs of identical letters in this word (M, A, and T). There are only 8 distinct letters. The
number of 3-letter strings that can be formed with 8 letters with no repeat letters is:
8 · 7 · 6 = 336. (To get this I multiplied the number of possibilities for the first letter, 8, by the number
of possibilities for the 2nd letter, 7 (all but the first letter), by the number of possibilities for the third
letter (all but the first 2 letters), 6. But we are allowed to have repeat letters up to two M's, or two A's, or
two T's. So how many strings are there with the first 2 letters the same? Well, we have a choice of 3
letters (M, A, or T) for the first letter. The second letter has to be the same as the first, so only 1 choice
there. The third letter can be any of the distinct letters except for the first letter, so 7 possibilities. So
there are 3 · 1 · 7 = 21 of these. Can also have strings where first and third letter match. Number of
these is 3 · 7 · 1 = 21. And, finally, can have strings where 2nd and 3rd letters match. Number of these
is 7 · 3 · 1 = 21. And that's it. So total number is 336 + 21 + 21 + 21 = 399.
Answer is B.