Combinations with Repetition – C(n+r−1, r)

When selection order does not matter and duplicates are welcome, the count is the multiset formula C(n+r−1, r). Choosing 3 scoops from 4 flavors gives C(6,3) = 20 possibilities — three vanilla, or two chocolate one strawberry, order never counted. Stars and bars makes it visual: r stars (picks) and n−1 bars (dividers between types) form a string of r+n−1 symbols; choosing which r positions hold stars — C(6,3) for our example — enumerates every multiset exactly once, e.g. ⭐|⭐⭐| | = one flavor-1 and two flavor-2 scoops. The calculator prints the formula substitution, the star-bar layout for small cases, and the contrast with plain combinations (C(4,3) = 4 without repeats). Distribution problems are the same count: 7 identical candies among 3 children split as C(9,2) = 36 arrangements.

1. 1

   Enter types n and picks r

   e.g. 4 flavors, 3 scoops

2. 2

   Apply C(n+r−1, r)

   C(6,3) = 20 with the substitution shown

3. 3

   View stars and bars

   Star/bar strings enumerate each multiset

4. 4

   Compare plain C(n,r)

   4 without repetition vs 20 with — repetition quintuples it

• 1

  Bars count n−1, not n or r

• 2

  Identical items into distinct bins uses the same formula

• 3

  Repetition always raises the count: 4 → 20 here

• Using plain C(n,r) and forgetting duplicate-friendly selections

• Treating order as mattering — that is nʳ, here 64

• Miscounting bars (n−1, not n)

• Excluding zero-pick types; empty bins are legal