Thread: Combinatorial argument and how many intersections?

1. Combinatorial argument and how many intersections?

1. Use a combinatorial argument to show that $\forall \ \{n,m,k\} \in \mathbb{Z^+}$ with $\{n,m\} \ge k$

$\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

2. We are given $\mbox{n}$ points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when $n = 6$there are $15$ intersections.

Thanks!

I just started combinatorics so please don't leave out any necessary steps! Thanks again

2. Originally Posted by usagi_killer
1. Use a combinatorial argument to show that $\forall \ \{n,m,k\} \in \mathbb{Z^+}$ with $\{n,m\} \ge k$

$\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

2. We are given $\mbox{n}$ points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when $n = 6$there are $15$ intersections.

Thanks!

I just started combinatorics so please don't leave out any necessary steps! Thanks again
do not post your questions in wrong subforums! this question is a "discrete math" question not abstract algebra.

3. Q1: How many ways can we choose k things from m+n things? Split the m+n bunch into a group of m things and a group of n things. Then you could choose the k things by choosing 0 from the n group and k from the m group. Or you could choose 1 thing from the n group and k-1 from the m group.
Or you could choose 2 things from the n group and k-2 from the m group.
etc.
How many ways are there of doing these? That's your sum.

Q2. Any 4 points determine 1 intersection point. Think 2 intersecting chords from 4 points. So you've determined n choose 4 points.