NIMCET 2012 Question Paper with solutions

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Question : 3 of 120

Marks: +1, -0

The sum of

↖{20}C_8

+

↖{20}C_9

+

↖{21}C_{10}

+

↖{22}C_{11}

-

↖{23}C_{11}

is

$↖{22}C_{12}$
$↖{23}C_{12}$
0
$↖{21}C_{10}$

Solution:

=

(\,↖{21}C_9+\,↖{21}C_{10})

+

↖{22}C_{11}

-

↖{23}C_{11}

; [

↖{n}C_r+\,↖{n}C_{r+1}

=

↖{n+1}C_{r+1}

]
=

(\,↖{22}C_{10}+\,↖{22}C_{11})

-

↖{23}C_{11}

=

↖{23}C_{11}-\,↖{23}C_{11}

= 0

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