Pascal's triangle and Pascal's rule
/ pas-KAL /
Write a 1 at the top. On each new row below, start and end with 1, and make every inner number the sum of the two numbers directly above it. You get a growing triangle: 1; then 1 1; then 1 2 1; then 1 3 3 1; then 1 4 6 4 1, and so on forever. This is Pascal's triangle, and its tidy arithmetic hides the entire family of binomial coefficients.
The numbers are not just pretty — the entry in row n, position k (counting from 0) is exactly C(n,k). So row 4 reading 1 4 6 4 1 is C(4,0), C(4,1), C(4,2), C(4,3), C(4,4). The 'add the two above' construction is a theorem in disguise, called Pascal's rule: C(n,k) = C(n-1, k-1) + C(n-1, k). Here is why it is true by pure counting: to choose k items from n, fix your eye on one particular item; either you include it (then choose the remaining k-1 from the other n-1, giving C(n-1,k-1)) or you exclude it (then choose all k from the other n-1, giving C(n-1,k)). Those two disjoint cases add up — the addition rule again.
Pascal's rule is the practical way to build binomial coefficients without factorials, just by adding. The triangle also displays the symmetry C(n,k) = C(n,n-k) as left-right mirror symmetry, and each row sums to 2^n. The same array appears centuries earlier in Chinese mathematics as Yang Hui's triangle, and it quietly encodes the binomial theorem, since row n lists the coefficients of (a+b)^n.
To get row 5 from row 4 (1 4 6 4 1), add neighbours: 1, (1+4)=5, (4+6)=10, (6+4)=10, (4+1)=5, 1, giving 1 5 10 10 5 1. Check against the formula: C(5,2) = 10. The sum 1+5+10+10+5+1 = 32 = 2^5.
Each entry is C(n,k); Pascal's rule C(n,k)=C(n-1,k-1)+C(n-1,k) builds the triangle.
Mind the indexing: rows and positions are usually counted from 0, so the single 1 at the top is row 0, and the second entry of row n is C(n,1) = n, not C(n,0).