Remove and replace letters in 'Arkansas' A sign reads 'Arkansas'. Three letters are removed and put back into the three empty spaces at random. What is the probability that the sign still reads 'Arkansas'?
My method
I tried a different case to get the feel of the problem.
I tried it for a smaller set of letters " M,I,T,A " and removed just two letters . For this case I found that the probability is 0.5. by taking cases. 
I still didn't get the feel of the problem. The repetition of A and S in Arkansas is bothering me. 
How should I approach this problem?
 A: If you pick three different letters, they have to be put back the same way for a chance of $1/3!$.  If you pick three the same they are guaranteed to make Arkansas.  If you pick two of one kind and one of another, what is the chance they go back to make Arkansas?  Now figure the chance of getting each type of selection.
A: Is "A" distitinct from "a"?  I would say it is.
You have two cases.  Either you pull 3 different letters. Or you pull 2 of 1 kind of letter, and one of the remainders. 
The second scenario is easier to count as inclusion exclusion comes into play in the first.
The doubled letter is either an "s" or an "a" and in each case there are 6 candidates for the companion letter.
12 was to choose two of one letter and 1 other.
${8\choose 3} = 56$ ways to choose 3 letters in all.
$56 - 12 = 44$ ways to choose 3 distinct letters.
If you really want a way to count the scenario of 3 distinct letters from "the bottom up"... ${4\choose 3} + 4{4\choose 2} + 4{4\choose 1} = 44$  one term assumes that no letters are from $\{a,s\}$ one term for choosing one letter from that set, and one term for choosing 2 letters from that set (but not two of the same variety)
If you choose 3 distinct letters, you have a $\frac 16$ of returning them from whence they came.
If you choose 2 distinct letters, you have a $\frac 13$ of returning them to equivalent positions.
A: Split it up into the cases where you pick letters that are non-repeating and the cases where you do pick letters that repeat (twice or thrice). 
Total # ways to pick any 3 letters from ARKANSAS = $^8C_3 = 56$ and the total ways to put 3 letters back = $3! = 6$. Therefore, total cases = $56*6 = 336$.
Number of ways to put back letters successfully when the chosen letters are all different = 1 and the number of possibilities where three different letters are chosen needs to be calculated. 
If A is picked as one letter, then there are $^5C_2$ ways of picking the other two letters such that they are not A's. 
However, out of these, there is 1 case where both of the other letters are S: ASS. So $^5C_2 - 1$ cases when A is one letter of three different ones. 
When S is one letter, $^6C_2$ ways to choose a non-S pair of letters, but out of which 3 cases will be AA's. So $^6C_2 - 3$. 
Therefore, # successes when three different letters are picked = $(^5C_2-1) + (^6C_2-3) + ^3C_3$ (using the non-repeating letters RKN)
Now for the cases that have repeating letters:
ASS can be put back correctly 2 ways. 
AAA can be put back correctly 3 ways.
AAX can be put back correctly 2 ways and X (non-A) can be chosen in 5 different ways.
AXA " ".
XAA " ".
Total = $(^5C_2-1) + (^6C_2-3) + ^3C_3 + 2 + 3 + 30 = 56$
Therefore, probability = $56/(56*6) = 1/6$
