Pick a or b from every bracket.

(a + b)ⁿ written as n brackets; tap a or b in each bracket to build one product, and all choices with r b's are tallied, giving nCr copies of aⁿ⁻ʳbʳ.

In (a + b)⁵ find all the ways to make a²b³ and confirm there are 10 = 5C3.

• The entries of Pascal's triangle are combinations: row n is nC0, nC1, nC2, …, nCn, where nCr = n!/(r!(n − r)!) for 0 ≤ r ≤ n and nC0 = nCn = 1.
• So any row can be written straight away. Row 7 is 7C0, 7C1, …, 7C7, that is 1, 7, 21, 35, 35, 21, 7, 1.
• The adding rule of the triangle is Pascal's rule from combinations: nCr + nC(r − 1) = (n + 1)Cr. The 1s at the ends are nC0 = nCn = 1.
• Each row is symmetric because nCr = nC(n − r). Row 6 reads 1, 6, 15, 20, 15, 6, 1 from either end.
• Why combinations appear: in (a + b)ⁿ written as n brackets multiplied together, a term aⁿ⁻ʳbʳ comes from picking b in r of the n brackets and a in the rest, and there are nCr ways to pick those r brackets.
• Counting paths gives the same numbers: on a board where a bead goes left or right at each of 6 rows of pins, the number of routes ending with r right-turns is 6Cr, so the 7 bins get 1, 6, 15, 20, 15, 6, 1 routes, 64 = 2⁶ in all.

Teaching this idea? Put the simulation on your own page or LMS with this code.