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Using Pascal's identity, $^7C_3 + ^7C_2$ equals:
A$^7C_5$, equal to the sum by symmetry of coefficients
B$^7C_4$, mistakenly using a different symmetry on Pascal
C$^6C_3$, mistakenly going down one row instead of up
D$^8C_3$, the next row's entry by Pascal's identity
Answer & Solution
Correct answer: D. $^8C_3$, the next row's entry by Pascal's identity
Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$, so $^7C_3 + ^7C_2 = ^8C_3$.
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