The sum $^{10} C_{3} +^{11} C_{3} +^{12} C_{3} + . . . . +^{20} C_{3}$ is equal to

^{21} C_{4}

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^{21} C_{4} -^{10} C_{4}

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^{21} C_{4} -^{11} C_{4}

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^{21} C_{17}

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Solution

The correct option is A $^{21} C_{4}$
$^{10} C_{3} +^{11} C_{3} +^{12} C_{3} + - - - +^{20} C_{3}$ ....(i)
addition and subtraction $^{10} C_{4}$ in equation (i)
$^{10} C_{4} -^{10} C_{4} +^{10} C_{3} +^{11} C_{3} + - - - +^{20} C_{3}$
[we know that - $^{n} c_{r - 1} +^{n} c_{r} =^{n + 1} c_{r}]$
$^{10} C_{3} +^{10} C_{4} +^{11} C_{3} + - - - +^{20} C_{3} -^{10} C_{4}$
$=^{11} C_{4} +^{11} C_{3} +^{12} C_{3} + - - - +^{20} C_{3} -^{10} C_{4}$
$=^{12} C_{4} +^{12} C_{3} + - - - +^{20} C_{3} -^{10} C_{4}$
$=^{13} C_{4} +^{14} C_{3} + - - - +^{20} C_{3} -^{10} C_{4}$
Similarly, Continue
So, we get $^{21} C_{4} -^{10} C_{4}$
So, option (B) is correct

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Similar questions

^{10} C_{3} +^{11} C_{3} +^{12} C_{3} + . . . . . . . . . . +^{20} C_{3}

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})

(21 C_{1} - 10 C_{1}) + (^{21} C_{2} - 10 C_{2}) + (21 C_{3} - 10 C_{3}) + (21 C_{4} - 10 C_{4}) + \dots . + (21 C_{10} - 10 c_{10})

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + . . . + (^{21} C_{10} -^{10} C_{10})

3

A, B

C

10, 12, 11

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