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# 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 domi

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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?

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Guest Mar 3, 2015

#1
+92206
+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.

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

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
Sort:

#1
+92206
+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.

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

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

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