alleryWhich identity is the basis for constructing Pascal's triangle row by row?
Aⁿ⁺¹C_r = ⁿC_r + ⁿC_(r−1)
BⁿC_r · ⁿC_(n−r) = 1
CⁿC_r = ⁿ⁻¹C_r
DⁿC_r = ⁿC_(r−1) · (n − r + 1)/r
Answer & Solution
Correct answer: A. ⁿ⁺¹C_r = ⁿC_r + ⁿC_(r−1)
The 'Pascal identity' ⁿ⁺¹C_r = ⁿC_r + ⁿC_(r−1) generates each entry by summing the two entries above it.
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