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Question

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

A
211111
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B
21116
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C
411111
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D
311111
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Solution

The correct option is B 21116
We can use binomial expansion to do this
(1+x)11=11C0+11C1x+11C2x2.........11C10x10+11C11x11
Now integrate the following code

10(1+x)11=11C0+11C1x+11C2x2.........11C10x10+11C11x11

(1+x)1212112=11C01+11C1x2+11C2x33.........11C10x1011+11C11x1212

Now put x=1

(1+1)12112=11C01+11C12+11C23.........11C1011+11C1112

(2)12112112=11C01+11C12+11C23.........11C1011

Thus

11C01+11C12+11C23.........11C1011=212212=21116

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