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Pascal's triangle is built from which recursion on combinations?
AC(n-1,k-1)+C(n-1,k)
BC(n-1,k) x C(n,1)
CC(n,k-1) - C(n,1)
D2 x C(n-1,k) + 1
Answer & Solution
Correct answer: A. C(n-1,k-1)+C(n-1,k)
Each entry is the sum of the two entries above it, which is the recursion for C(n, k).
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