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A grid with $3$ rows and $52$ columns is tiled with $78$ identical $2\cdot1$ dominoes. In how many ways can this be done such that exactly two of the dominoes are vertical.
I tried-
Both the dominoes would be in the same row. And they should be spaced in such a way that more dominoes can be placed between them. So for the first place, there are $26$ choices. For second, $25$, and so on. So $26+25+24+23+22+21+...+1=\frac{26.27}{2}$
Now same thing would apply for second and third row, so result would be doubled, which gives $\frac{26.27}{2}.2=702$

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  • $\begingroup$ Can you explain the answer you got? What does $\binom{78}2$ come from? I guess that means you're picking $2$ of the $78$ dominoes? But what's the point of that if the dominoes are identical? $\endgroup$ – bof Apr 19 '15 at 13:41
  • $\begingroup$ I know its wrong but also I have no other idea. $\endgroup$ – Aditya Agarwal Apr 19 '15 at 13:56
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Number the three rows $1,2,3$ from top to bottom. Suppose we put a vertical domino so it covers a square in row $1$ and a square in row $2$. How many squares are left uncovered in row $1$? Can we cover that exact number of squares if we use the other vertical domino to cover one of them? Can we cover that exact number of squares if we do not use the vertical domino, that is, we use only horizontal dominoes?

If you actually draw some $3\times n$ grids where $n$ is an even number, for example $n=8$, and try drawing the dominoes that would cover them (or perhaps better, find some actual dominoes or domino-shaped objects and try arranging them in rectangles of height $3$ and even width), you may notice some patterns to the way the horizontal dominoes are placed when they are in the same rows as a vertical domino.

Suppose for example that you place two horizontal dominos in rows $1$ and $2$, overlapping like this:

enter image description here

In which rows can you put a vertical domino then? Try it.

In fact, once you have decided where one of the vertical dominoes goes, I think you may notice that you don't have a huge number of choices remaining about how to place the rest of the dominoes.

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  • $\begingroup$ I am getting the answer- $4.(26+\frac{17.18}{2})$ $\endgroup$ – Aditya Agarwal Apr 20 '15 at 9:48
  • $\begingroup$ 716? Is it right? $\endgroup$ – Aditya Agarwal Apr 20 '15 at 9:49
  • $\begingroup$ Closer. You seem to be double-counting something, but it's hard to say what since you never say how you get any of your numbers. $\endgroup$ – David K Apr 20 '15 at 12:45
  • $\begingroup$ Sorry. Now I got it. The dominoes should be in the same row. So $2(\frac{26.27}{2})$. Because in the same row, there are $26+25+...+1$ ways and then for both there will be double the ways. So $702$. Now if we rotate the dominoes or reverse them, then there will be $702*2=1404$ ways. $\endgroup$ – Aditya Agarwal Apr 20 '15 at 15:19
  • $\begingroup$ This looks just like the way I count it, except that I don't understand the "rotate or reverse" step (so I only counted $702$ ways). $\endgroup$ – David K Apr 20 '15 at 16:54
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Hint 1: In each row, horizontal dominoes cover an even number of squares in the row.

Hint 2: In a row intersected by vertical domino(es), the remaining squares have to come in even numbered runs so that they can be covered by horizontal dominoes.

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  • $\begingroup$ Please tell the solution. I just cannot make out the solution. $\endgroup$ – Aditya Agarwal Apr 19 '15 at 13:56
  • $\begingroup$ Maybe try a few smaller grids by hand--such as $3\times 4$, $3\times 6$, $3 \times 8$, and $3\times 10$; and see what you learn from these small cases, keeping the hints in mind. $\endgroup$ – paw88789 Apr 19 '15 at 14:28

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