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Math Challenge II-A Combinatorics

 
 
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Math Challenge II-A Combinatorics
by Neo Liang - Tuesday, September 3, 2019, 5:08 PM
 

Lecture 2, Problem 2.25

I got $210$, but the answer says that it's $245$. There are $\dbinom{6}{2} \cdot 7$ squares because there are $\dbinom{6}{2}$ combinations for the horizontal sides and $\frac{7 \cdot 2}{2}$ combinations for the vertical sides. There are $\dbinom{7}{2}\dbinom{6}{2}$ rectangles, so the answer is $\dbinom{7}{2}\dbinom{6}{2}-\dbinom{6}{2} \cdot 7$, or $\dbinom{6}{2}\left(\dbinom{7}{2}-7\right)$, or $15 \cdot 14$, which is $210$.

 
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Re: Math Challenge II-A Combinatorics
by Areteem Professor - Wednesday, September 4, 2019, 1:32 PM
 

Can you explain a little more where the $\binom{6}{2}\cdot 7$ answer comes from for the number of squares? (The answer given uses cases to count the number of squares.)

You are correct that there are $\binom{6}{2}$ ways to choose a collection of $2$ lines for the horizontal sides of the square. However, depending on which lines are chosen, there are varying numbers of squares chosen. For example, if the two horizontal lines are $1$ apart, there are $6$ possible ($1\times 1$) squares. However, if we choose the top and bottom horizontal lines, there are only $2$ possible ($5\times 5$) squares. This is why the different cases are needed.