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

$2^{21} - 2^{10}$

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$2^{20} - 2^{9}$

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$2^{20} - 2^{10}$

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$2^{21} - 2^{11}$

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Solution

The correct option is C
$2^{20} - 2^{10}$

Explanation for the correct answer:
On rearranging the terms of the given expression we get,
$({}^{21}{C_{1} - {}^{10}C_{1}}) + ({}^{21}{C_{2} - {}^{10}{C_{2}}}) + ({}^{21}{C_{3} - {}^{10}{C_{3}}}) + . . . . . . + ({}^{21}{C_{10} - {}^{10}{C_{10}}}) = ({}^{21}{C_{1} + {}^{21}{C_{2}} + . . . . . + {}^{21}{C_{10}}}) - ({}^{10}{C_{1}} + {}^{10}{C_{2}} + . . . . + {}^{10}{C_{10}})$ $. . . (i)$
We know, the sum of ${}^{n}{C_{1} + {}^{n}{C_{2}} + {}^{n}{C_{3}} + . . . . . + {}^{n}{C_{n}}} = 2^{n} - 1$
$\Rightarrow$ ${}^{10}{C_{1} + {}^{10}{C_{2}} + {}^{10}{C_{3}} + . . . . . + {}^{10}{C_{10}}} = 2^{10} - 1$ $. . . (i i)$
$({}^{21}{C_{1} + {}^{21}{C_{2}} + . . . . . + {}^{21}{C_{10}}}) = \frac{1}{2} (({}^{21}{C_{1} \times 2 + {}^{21}{C_{2} \times 2} + . . . . . + {}^{21}{C_{10}}} \times 2))$
We know that, ${}^{n}{C_{r} = {}^{n}{C_{n - r}}}$
$\Rightarrow$ ${}^{21}{C_{1} = {}^{21}{C_{20,}}} {}^{21}{C_{2} = {}^{21}{C_{19}}, . . . . . ., {}^{21}{C_{10}} = {}^{21}{C_{11}}}$
$\Rightarrow$ $\frac{1}{2} (({}^{21}{C_{1} \times 2 + {}^{21}{C_{2} \times 2} + . . . . . + {}^{21}{C_{10}}} \times 2)) = \frac{1}{2} ({}^{21}{C_{1} + {}^{21}{C_{2}}} + . . . + {}^{21}{C_{10}} + {}^{21}{C_{11} + {}^{21}{C_{12}}} + . . . . + {}^{21}{C_{20}})$
$= \frac{1}{2} ({}^{21}{C_{1} + {}^{21}{C_{2}}} + . . . + {}^{21}{C_{10}} + {}^{21}{C_{11} + {}^{21}{C_{12}}} + . . . . + {}^{21}{C_{20} + {}^{21}{C_{21}}} - {}^{21}{C_{21}})$
$= \frac{1}{2} (2^{21} - 1 - \frac{21!}{21!})$ $(∵ {}^{21}{C_{1} + {}^{21}{C_{2} +} . . . . . {}^{21}{C_{21}}} = 2^{21} - 1)$
$= \frac{1}{2} (2^{21} - 2)$
$= 2^{20} - 1$ $. . . (i i i)$
From $(i), (i i), (i i i)$ we get,
$({}^{21}{C_{1} - {}^{10}C_{1}}) + ({}^{21}{C_{2} - {}^{10}{C_{2}}}) + ({}^{21}{C_{3} - {}^{10}{C_{3}}}) + ({}^{21}{C_{4} - {}^{10}{C_{4}}}) + . . . . . . + ({}^{21}{C_{10} - {}^{10}{C_{10}}}) = 2^{20} - 1 - 2^{10} + 1$
$= 2^{20} - 2^{10}$
Hence, option $(C)$ , $2^{20} - 2^{10}$ is the correct answer.

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