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Question

11C01+11C12+11C23+....+11C1011=

A
211111
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B
212112
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C
311111
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D
31116
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Solution

The correct option is B 212112
We know,
(1+x)11=11C0+11C1x+11C2x2+....+11C11x11
Integrating both sides with respect to x:
(1+x)1212=11C0x1+11C1x22+11C2x33+....+11C11x1212
Putting x=1
(2)1212=11C01+11C12+11C23+....+11C1112

11C01+11C12+11C23+....+11C1011=21212112=212112

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