How this instrument works
Pascal's Triangle is built by addition alone. Row 0 holds a single 1; every later row starts and ends with 1, and each entry in between is the sum of the two entries diagonally above it in the row before. Carry that rule out six rows and it produces 1, 5, 10, 10, 5, 1 — and every number the rule produces is also a binomial coefficient, entry(n,k) = C(n,k), because building a set of k items from n by first deciding on one particular item either includes it or leaves it out, and those two cases are exactly the two entries the addition rule sums.
Shade every odd number in the triangle and leave the even ones blank, and a fractal appears: the Sierpinski triangle, the same nested pattern of triangular gaps that turns up in fractal geometry with no combinatorics in sight. The reason is Lucas' theorem: an entry is odd exactly when every binary digit of its position k is already switched on in the binary digits of its row n, a parity rule that repeats itself at every doubling of scale and produces the self-similar holes.
A neat limit sits at the small end of the triangle. Read each of the first five rows as the digits of one number and it equals a power of eleven exactly — row 0 is 1, row 1 is 11, row 2 is 121, row 3 is 1331, row 4 is 14641. Row 5 is where it breaks: the rule needs a two-digit entry, 10, so simply writing 1 5 10 10 5 1 side by side no longer matches 11^5 = 161051 without carrying the extra digit the way long multiplication would.
- Enter the row to read in Row, n, counting from 0 at the triangle's apex.
- Enter the spot along that row in Position, k, also counting from 0 at the row's left-hand 1.
- Read Entry: C(n, k) for the number sitting at that row and position.
- Keep k at or below n — position 6 does not exist in a six-wide row 5, and the sheet flags any k greater than n.
- Step n up or down a row at a time and watch each entry track the sum of the two entries above it.
Worked example — row 5, position 2
Set Row, n to 5 and Position, k to 2. Row 4 of the triangle is 1, 4, 6, 4, 1, so the addition rule builds row 5 as 1, 5, 10, 10, 5, 1 — and position 2, counting from 0, is 10. The factorial formula agrees exactly: C(5,2) = 5! ⁄ (2!·3!) = 120 ⁄ (2×6) = 120 ⁄ 12 = 10.
The addition rule checks the same answer a second way: entry(5,2) should equal entry(4,1) + entry(4,2). Row 4 gives entry(4,1) = 4 and entry(4,2) = 6, and 4 + 6 = 10, matching the factorial route digit for digit. Both paths land on 10, which is the value this sheet returns for n = 5, k = 2.
Questions
How do you build Pascal's Triangle without any formula?
Start with a single 1 as row 0. Every later row begins and ends with 1, and each number between those two is the sum of the pair sitting diagonally above it in the previous row. Row 1 is 1, 1; row 2 is 1, 2, 1; row 3 is 1, 3, 3, 1 — no factorials needed, just repeated addition.
Why does shading the odd numbers turn the triangle into a fractal?
Because parity in the triangle follows Lucas' theorem: entry(n,k) is odd only when every binary digit of k is already present in the binary digits of n. That rule repeats identically at each doubling of scale, so the pattern of odd (shaded) and even (blank) cells nests inside itself and produces the Sierpinski triangle.
Does a row of the triangle really equal a power of eleven?
Only through row 4. Reading the digits of rows 0 to 4 side by side gives 1, 11, 121, 1331, and 14641 — exactly 11^0 through 11^4. Row 5 needs a two-digit entry, 10, so the shortcut breaks there; the underlying link still holds if you carry the extra digit the way long multiplication would, but the simple digit-reading trick stops.
What happens if Position, k is set higher than Row, n?
There is no entry there — a row of width n+1 only has positions 0 through n, so k=n+1 falls off the end. This sheet's check catches it directly: k cannot exceed n, and it flags the input rather than guessing at a value outside the row.
Is there anything meaningful along the triangle's diagonals?
Yes. The first diagonal is all 1s, the second is the counting numbers 1, 2, 3, 4…, and the third is the triangular numbers 1, 3, 6, 10… Sum along the shallow diagonals that cut across the rows at a slant and the totals are the Fibonacci numbers, 1, 1, 2, 3, 5, 8… — a pattern with no obvious link to counting subsets.