*Restriction: Once the first letter is picked, the second letter must be to the right of the first letter in the word "MATHEMATICS". For example, if we were to pick "S" as the first letter, there is no way to pick the second letter because there is not alphabet to the right of "S".
My attempted answer: 43.
My reasoning:
1. label "MATHEMATICS" with numbers (1,2,3,…,11) correspondingly (1 for "M", 2 for "A", 3 for "T", 4 for "E", etc.)
2. Write our choice of two alphabets as (a,b). Note, a choice (a,b) is valid only when b>a, where a,b $\in${1,2,…,11}.
3. Plot valid (a,b) as dots in cartesian coordinate. Count the number of valid (a,b)'s , where b>a. The number of such (a,b)'s should be $\frac{11^2-11}{2}$ (geometrically we are counting the number of dots above the left to right diagonal in a 11 by 11 square). There are 55 of such (a,b)'s.
4. Assumption: I consider "MA" (1,2) the same word pick as "MA" (6,7). Now to know how many distinct pairs of two-alphabet pairs (a,b) there are, we subtract the number of repeats. E.g. "MA" (1,2) and "MA" (6,7) are counted twice in the 55 pairs of (a,b). They are repeats.
5. The number of repeats are (11-6)+(11-7)+(11-8)=12. We can see this once we draw (a,b)'s as points in Cartesian coordinate.
6. With my reasoning, the final answer should be 55-12=43. However, I only found 40 of those distinct pairs in all.
My reasoning must have been wrong. Please help me out. Ps. I asked the question without *Restriction, while I had meant to ask with it.
Best Answer
You need to pick different letters in each sequence, so the alphabet "MATHEMATICS" has "MAT" repeated, so you have $8$ distinct letters.
Now distribute two letters having $8$ letters in total:
$\binom{8}{2} = \frac{8!}{2!6!} = 28$ those are combinations, but you need to permute them in order to include all the pairs, so:
$28.2 = 56$
EDIT: In the case you consider all the letters from "MATHEMATICS" distinct, then you have 11 distinct letters therefore when you pick one letter, the second one cannot be any letter from the left.
MATHEMATICS $= \{1,2,3,4,5,6,7,8,9,10,11\}$
When you pick $11$ there's no possibility since rest of letters are in the left. With $10$ you have 1 possibility, $11$ etc. until you reach $1$ that has $10$ possibilities.
This is summarized as following: $\sum_{i=1}^{10}\binom{i}{1}=55$
with alphabet $\sum = \{1,2,3,4,5,1,2,3,9,10,11\}$
$10=1,9=2,3=5,2=6,1=7,5=3,4=4,3=5,2=6,1=7 \Rightarrow 1+2+5+6+7+3+4+5+6+7 = 46$