How can you prove the N choose k formula without using a formula?

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The discussion focuses on proving the combinatorial identity for m, n ≥ 2: (m+n choose 2) = (m choose 2) + (n choose 2) + mn. The first proof utilizes the formula for combinations, specifically a(n choose k) = a! / ((a - k)!k!). The second proof, which avoids using this formula, suggests employing Pascal's triangle and counting 2-subsets from two disjoint sets with m and n elements. This approach highlights the relationship between combinatorial counting and arithmetic series.

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mattmns
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Hello.

I have the following problem: Show that for [tex]m,n \geq2[/tex],

[tex]\left(\begin{array}{cc}m+n\\2\end{array}\right) = \left(\begin{array}{cc}m\\2\end{array}\right) + \left(\begin{array}{cc}n\\2\end{array}\right) + mn[/tex]

by using the formula

[tex]\left(\begin{array}{cc}a\\b\end{array}\right) = \frac{a!}{(a - b)!b!}[/tex]

and algebra. Prove it again without using this formula.

The first part was quite easy, but I am not sure how I could solve the second (bold) part without using a formula. Am I supposed to use a definition or something of that nature? I just am not seeing it, any ideas would be appreciated. Thanks
 
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You could use Pascal's triangle and note that the third item in row N is the sum of the items in the third diagonal with values [itex]1+2+ \cdot\cdot\cdot \+ N-1[/itex] which is just an arithmetic series.
 
Take two disjoint sets, one having m elements and the other having n elements, and count the number of 2-subsets of their union in two different ways.
 

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