Practice problem
What does the horizontal symmetry of Pascal's Triangle tell us about combinatorics?
For any positive iii, (2ii)≠(nk){2i \choose i} \not= {n \choose k}(i2i)=(kn) for n≠2in \not= 2in=2i, k≠ik \not= ik=i.
(nn)=1{n \choose n} = 1(nn)=1
(nk)=(nn−k){n \choose k} = {n \choose n-k}(kn)=(n−kn)
(nk)=(nn−k+1){n \choose k} = {n \choose n-k+1}(kn)=(n−k+1n)
(nk)=(n2k){n \choose k} = {n \choose 2k}(kn)=(2kn)
Type a concept, like “slope” or “photosynthesis”.