The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

2^{21} - 2^{11}

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2^{21} - 2^{10}

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2^{20} - 2^{9}

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2^{20} - 2^{10}

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Solution

The correct option is D $2^{20} - 2^{10}$
$(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + \dots + (^{21} C_{10} -^{10} C_{10})$
$= (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10}) - (^{10} C_{1} +^{10} C_{2} + \dots +^{10} C_{10})$

$Now,^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10}$
$= \frac{1}{2} (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10} +^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10})$
$= \frac{1}{2} (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10} +^{21} C_{20} +^{21} C_{19} + \dots +^{21} C_{11})$
$= \frac{1}{2} (2^{21} - 2) = 2^{20} - 1$

$^{10} C_{1} +^{10} C_{2} + \dots +^{10} C_{10}$
$= 2^{10} - 1$

$∴$ Required sum $= 2^{20} - 2^{10}$

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Similar questions

(^{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^{1} \times^{10} C_{1} + 3^{2} \times^{10} C_{2} + 3^{3} \times^{10} C_{3} + . . . . + 3^{10} \times^{10} C_{10} =

^{10} C_{1} +^{10} C_{2} +^{10} C_{3} + . . . +^{10} C_{9}

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