Online Course Discussion Forum

Math Challenge II-A Combinatorics

 
 
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Re: Math Challenge II-A Combinatorics
by Areteem Professor - Wednesday, 4 September 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.