*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.

  • $\begingroup$ Thank you all for answers. But if I were to impose a new rule such that once you pick the first letter, the second letter cannot be any letter that comes before the first letter (in this case, "CT" would be an invalid pick because in "MATHEMATICS" you cannot choose a C s.t. you find a T that comes after "C"), then is my reasoning posted in the question valid? With this restrict, if we were to consider "MA" (1,2) indistinguishable from "MA"(6,7), then a solution based on my reasoning yields an answer 55-(5+4+3)=43. $\endgroup$ – HenHar Dec 11 '16 at 19:39
  • $\begingroup$ I've edited my answer providing more information to your last answer. $\endgroup$ – kub0x Dec 11 '16 at 20:04

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 $\sum = \{1,2,3,4,5,1,2,3,9,10,11\}$

$10=1,9=2,3=5,2=6,1=7,5=3,4=4,3=5,2=6,1=7 \Rightarrow 1+2+5+6+7+3+4+5+6+7 = 46$

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  • $\begingroup$ @kubox. Thank you for your answer. I added a restriction to the question in the comment. Could you please take a look? $\endgroup$ – HenHar Dec 11 '16 at 19:46
  • $\begingroup$ Thanks! If we consider "MA" (1,2) different from "MA"(6,7), i.e., MATHEMATICS:=(1,2,3,4,5,1,2,3,9,10,11) we just subtract the repeats, and there are (11-6)+(11-7)+(11-8) of them. So 55-12=43 should be the answer? $\endgroup$ – HenHar Dec 11 '16 at 20:14
  • $\begingroup$ I'm gonna edit my answer with alphabet = (1,2,3,4,5,1,2,3,9,10,11) $\endgroup$ – kub0x Dec 11 '16 at 20:24
  • $\begingroup$ You double counted on 1,2,3, and counted wrong on each number (e.g. with your way of counting, the first 1 corresponds to 8 possible choices of the second number, and the second 1 should correspond to 5). I will try to count all possibilities again since I only found fewer than 43 possibilities... $\endgroup$ – HenHar Dec 11 '16 at 21:09

The distinct letters in the word MATHEMATICS are 8:M,A,T,H,E,I,C,S. Now, to choose two different letters, we have to pick any two letters from this eight lettered set. If order of picking is important,it is given by $8P2=\frac{8!}{6!}=56$

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  • $\begingroup$ Thank you for your answer. I added a restriction to the question in the comment. Could you please take a look? $\endgroup$ – HenHar Dec 11 '16 at 19:45

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