A grid with 3 rows and 52 columnsmins tiled with 78 identical 2x1 dominoes. How many ways can this be done such that exactly two of the dominoes are vertical?

Guest Mar 3, 2015

#1**+5 **

Well, The 2 vertical ones will have to both be in rows 1 and 2 OR both be in rows 2 and 3.

I am going to look at how many ways they can be in rows 1 and 2 (the number of positions in rows 2 and 3 will be the same)

There will be 25 horizonal tiles in each of rows 1 and 2 and these tiles must be exactly under each other. they can not be interlaced like bricks , or they won't all fit.

There must be 2 vertical ones placed between them somewhere, anywhere.

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I am going to look at an easier problem to get an understanding of this. Say there were just 3 horzontal bricks, how many places could the vertical bricks be placed?

xx [brick] xx [brick] xx [brick] xx

the crosses are where the vertial ones could go. There are 2(3+1) = 8 posibilities. We need just 2 of these.

So I think that there are 8C2 posibilities

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BACK to your problem 2(25+1)=100 possibilities We need just 2 of these.

So I think that there are 100C2 possible places for the 2 vertical bricks. This is just for rows 1 and 2.

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Altogether I think that there are 2*100C2 arrangements possible.

$${\mathtt{2}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{100}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{9\,900}}$$

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I think that is right but I won't stake my life on it. LOL

Melody Mar 4, 2015

#1**+5 **

Best Answer

Well, The 2 vertical ones will have to both be in rows 1 and 2 OR both be in rows 2 and 3.

I am going to look at how many ways they can be in rows 1 and 2 (the number of positions in rows 2 and 3 will be the same)

There will be 25 horizonal tiles in each of rows 1 and 2 and these tiles must be exactly under each other. they can not be interlaced like bricks , or they won't all fit.

There must be 2 vertical ones placed between them somewhere, anywhere.

------------------------

I am going to look at an easier problem to get an understanding of this. Say there were just 3 horzontal bricks, how many places could the vertical bricks be placed?

xx [brick] xx [brick] xx [brick] xx

the crosses are where the vertial ones could go. There are 2(3+1) = 8 posibilities. We need just 2 of these.

So I think that there are 8C2 posibilities

--------------------------

BACK to your problem 2(25+1)=100 possibilities We need just 2 of these.

So I think that there are 100C2 possible places for the 2 vertical bricks. This is just for rows 1 and 2.

---------------------------

Altogether I think that there are 2*100C2 arrangements possible.

$${\mathtt{2}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{100}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{9\,900}}$$

----------------------------

I think that is right but I won't stake my life on it. LOL

Melody Mar 4, 2015