A supermove in Freecell is when you move more than one card at the same time and automatically takes into account any free cells and free columns you have in order to maximise the number you can move. The exact formula is not a problem: 2^emptycolumns * (freecells+1)  My question is, when I make a supermove how can I calculate how many moves were made to cells. I am not talking about how many cells were used. This would be easy to calculate. What I mean is, for example, you move 10 cards because you have 4 free cells and 3 free columns, it is simulating a number of steps that include moving to cells and columns and back. So, how many moves were to cells?  Edit:  does not answer my question. That just gives the formula to calculate how many cards can be moved but not how many cell moves are needed.
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I'm working on a FreeCell implementation and I'm currently stuck determining how many cards the user is allowed to move from tableau to tableau at once. I've seen multiple formulas for this, but no explanation them. (empty_columns is decreased by 1 if we move to an empty column)   2 ^ empty_columns + free_cells 2 ^ empty_columns * (free_cells+1) (empty_columns+1) * (free_cells+1)   What I'd like to know is if one of these is correct or (even better) anyone knows a small algorithm to move cards like the in Microsoft implementation  I'm asking this in this stack because this question is not about the implementation itself, but about the game mechanics.
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How to use the  method?  Say I want to find number of combinations I can get with $x_1+x_2+x_3+x_4=22$,  where $x_i\in\mathbb{N}$.  Is this the correct time to apply the method?  This is being repurposed in an effort to cut down on duplicates, see here:     and here:
110540
The multiple-choice question:  A hotel has four vacant rooms. Each room can accommodate a maximum of four people. In how many different ways can six people be accommodated in the four rooms? (A) 4020 (B) 4068 (C) 4080 (D) 4096.  My attempt:  If there were no 'four person' constraint, the six people could be accommodated in $4^6=4096$ ways. however, this number includes combinations with 5 or 6 people in one room, which should be excluded.  If one room has 5 occupants, one other room has 1 occupant. The number of '1,5' combinations is ${{4} \choose {2}}$=12. If one room has 6 occupants, the other three rooms have no occupants, and there are four such combinations.  I believe the answer should be $4096-12-4=4080$, but am not confident. I'd be grateful for any comments on my attempt.
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I have been reading about Combinatorial Material for a while now.  I also solved a few examples. However am stuck at this one.     Find the total number of ways a $3 \times n$ board can be painted using $3$ colors while making sure no cells of the same row or the same column have entirely the same color.  The answer must be computed modulo $10^9 + 7$.   I saw the solutions to $3 \times n$ using  colors but all these had the constraint that no adjacent cells were to have the same color. There's no such constraint here.  So am just not sure how to do this given the center row can have a bunch of options as well.  Thanks for any help.
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I'm using supertabular in multicols enviroment using this trick  \let\mcnewpage=\newpage \makeatletter \renewcommand\newpage{%     \if@firstcolumn         \hrule width\linewidth height0pt         \columnbreak     \else         \mcnewpage     \fi } \makeatother   I'd like to put that code into \newcommand - I've tried this:  \newcommand{\TrickSupertabularIntoMulticols}{ \let\mcnewpage=\newpage \makeatletter \renewcommand\newpage{%     \if@firstcolumn         \hrule width\linewidth height0pt         \columnbreak     \else         \mcnewpage     \fi } \makeatother }   but when i use my \newcommand it doesn't work - table is in only one column. What is the right way to define this command?  EDIT: examples  this works fine but i generate multiple multicols in the document so I want to put that trick code into single line comand   \documentclass{article} \usepackage{supertabular,multicol}  \newcount\n \n=0 \def\tablebody{} \makeatletter \loop\ifnum\n&lt;100         \advance\n by1         \protected@edef\tablebody{\tablebody                 \textbf{\number\n.}&amp; shortText                 \tabularnewline         } \repeat \makeatother  \begin{document} \begin{multicols*}{2} \let\mcnewpage=\newpage \makeatletter \renewcommand\newpage{%         \if@firstcolumn                 \hrule width\linewidth height0pt                 \columnbreak         \else                 \mcnewpage         \fi } \makeatother  \begin{supertabular}{|l|l|} \tablebody \end{supertabular} \end{multicols*} \end{document}   and this doesn't work  \documentclass{article} \usepackage{supertabular,multicol}  \newcount\n \n=0 \def\tablebody{} \makeatletter \loop\ifnum\n&lt;100         \advance\n by1         \protected@edef\tablebody{\tablebody                 \textbf{\number\n.}&amp; shortText                 \tabularnewline         } \repeat \makeatother  \newcommand{\TrickSupertabularIntoMulticols}{ \let\mcnewpage=\newpage \makeatletter \renewcommand\newpage{%         \if@firstcolumn                 \hrule width\linewidth height0pt                 \columnbreak         \else                 \mcnewpage         \fi } \makeatother }  \begin{document} \begin{multicols*}{2} \TrickSupertabularIntoMulticols \begin{supertabular}{|l|l|} \tablebody \end{supertabular} \end{multicols*} \end{document}
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After playing a little with a Rubik cube I thought of the following problem :      Suppose we start with a solved Rubik cube (a general one , with $n^3$ cubes) . Then we choose one of the moves , each having a probability of $\frac{1}{6n}$ of being chosen , and apply it on our cube . We continue doing so ( choosing a new move randomly and then applying it and so on...) , until we reach the solved position again . What is the expected number of moves needed to solve the cube in this way ?   I thought about it and I think the answer should be $\infty$ (at least for $n \geq 3$) because there are a lot of random sequences of moves that will take a long time to solve the cube but I don't have a rigorous way to prove it.  For $n=2$ I'm not that sure if the answer should be finite or infinite (but I still tend to think it's infinite).  Thank you for your time to help me!
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I have the following problem from book "Introduction to Probability", p.32     A certain casino uses 10 standard decks of cards mixed together into one big deck, which   we will call a superdeck. Thus, the superdeck has 52 · 10 = 520 cards, with 10 copies   of each card. How many different 10-card hands can be dealt from the superdeck? The   order of the cards does not matter, nor does it matter which of the original 10 decks   the cards came from. Express your answer as a binomial coefficient.   Hint: Bose-Einstein.   My solution:  Because the number of cards of each type in the superdeck (10) is not less than the size of the hand (10), and thus not limiting, it's the same as sampling with replacement where the order does not matter,  so the number of possible 10-card hands would be $\binom{52+10-1}{10}$.  Is my thinking correct?
59085
I came across an interesting puzzle:     You are climbing a stair case. It takes $n$ steps to reach to the top.   Each time you can either climb $1$ or $2$ steps. In how many distinct   ways can you climb to the top?   Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?
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I was looking at this website  and I read the explanation for how to calculate the probability of getting a full house.  To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.  In other words, for a full house $P=$ $$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$  Following this logic, I tried to calculate the probability of getting two pair.  My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.    So I tried $P=$ $$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$  Obviously my solution was incorrect.  I read explanation and the correct answer is $P=$ $$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$  I'm still a bit fuzzy on where I went wrong though.  Can anyone help me understand this problem a little better?  Thank you very much for your help.
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This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer.  What i want to know is why my solution is not equivalent to the actual solution.  Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.  So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)  The logic is the following:  1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.  2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.  Total combinations: 54912  My solution is the following:  ((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))  1) My logic for the 3 of a kind is the same as the solution.   2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.  3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.  Total combinations: 109824  My solution says that the combinations are exactly double of the book's solution.  I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you.
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I saw this question but I couldn't find the answer. Assume that we have a 10x10 table, and it's filled with 0 to 9 numbers ( 10 of each of them are in the table, 10x zero, 10x one, and ... ) By using graph, prove that we have a column or a row that we can find at least 4 different numbers in it. edit: I have found this method to create a graph from any table-matrix :  after that, I thought that it would be a good way to use coloured edges. and now the thing to prove is to have a vertex with at least four colour edges connected to it. maybe with pigeonhole, the problem can be solved but I couldn't
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I don't understand why a set with n elements has 2^n subsets. How is this calculated? I realize that {123} has empty set - 1-2-3-1,2-1,3-2,3-1,2,3 but how is the formula derived?
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What is the distance a creature or player can travel if they have multiple types of movement?  For example, the  has a 40 ft walking speed, a 40 ft climbing speed, and an 80 ft flying speed. Can it move a total of 160 ft in a round using each of those fully, or is there another way this is calculated?
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Yesterday my housemates and I were playing card games and someone popped this question. We tried to solve the problem, but we couldn't figure it out. This morning I woke up and I am still wondering how to solve it. Will you please help me out?
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A 5x5 board has $25$ cells. The numbers $\{1,2,3,4,5\}$ are written on every row,every column and the two main diagonals without any repetition. If the sum of the numbers of the diagonal below the leading diagonal be represented as $A$. Show that $A\ne 20$. Also find the maximum possible value of $A$.    I could prove that $A\ne 20$ by showing that each of the cells of the diagonal below the leading diagonal cannot be $20$ as if such arrangement exists, then the board cannot have $5$ five times, which actually should be by the given condition of the question.    For the second part, I can imagine the maxiumum value to be $17$. For which I proved that $5$ can occur only $2$ times in the four cells and probably $4$ cannot occur twice. But I got no formal proof. Please help!
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I would like to start a new excel spreadsheet, create a column named Column1, and quickly fill in increments, (Column2, Column3 etc) up to Column500. Is there an easy way to do this?   Also is there an easy way to afterwards select all 500 columns and add 10 rows of data to all of them?
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How do I automatically average the lowest 3 of the last 6 numbers in a row of 22 that adds one number per week?  Obviously I need at least 7 numbers to begin using this formula.  Also, some cells in some rows may be blank.