The number of distinct terms in the expansion of (x+y2)13+(x2+y)14 is
A
27
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B
29
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C
28
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D
25
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Solution
The correct option is C 28 To get common terms in both the expansions xr1(y2)13−r1=(x2)r2(y)14−r2 r1=2r2 & 26−2r1=14−r2 r1=8;r2=4 ∴Only one term is common.