JEE Main 2019 — Mathematics Algebra

There is only one way in which we can choose one, two or more objects from $10$10 identical objects.

Number of ways of selecting $10$10 objects

$$={ }^{21}{C}_{10}+{ }^{21}{C}_{9}+{ }^{21}{C}_{8}+{ }^{21}{C}_{7}+{ }^{21}{C}_{6}+{ }^{21}{C}_{5}+{ }^{21}{C}_{4}+{ }^{21}{C}_{3}+{ }^{21}{C}_{2}+{ }^{21}{C}_{1}+{ }^{21}{C}_{0}$$=21C10​+21C9​+21C8​+21C7​+21C6​+21C5​+21C4​+21C3​+21C2​+21C1​+21C0​

We know ${ }^{n}{C}_{r}={ }^{n}{C}_{n-r}$nCr​=nCn−r​ and

\Rightarrow {2}^{21}={ }^{21}{C}_{0}+{ }^{21}{C}_{1}+{ }^{21}{C}_{2}+{ }^{21}{C}_{3}+{ }^{21}{C}_{4}+.........+{ }^{21}{C}_{21} {\begin{matrix}\because { }^{21}{C}_{21}={ }^{21}{C}_{0} \\ { }^{21}{C}_{20}={ }^{21}{C}_{1} \\ { }^{21}{C}_{19}={ }^{21}{C}_{2}\end{matrix}

$$=2({ }^{21}{C}_{0} +{ }^{21}{C}_{1} +{ }^{21}{C}_{2} +\ldots \ldots \ldots \ldots \ldots +{ }^{21}{C}_{10})$$=2(21C0​+21C1​+21C2​+……………+21C10​)

$$\Rightarrow { }^{21}{C}_{0}+{ }^{21}{C}_{1} {+}^{21}{C}_{2} +.........+{ }^{21}{C}_{10} = {2}^{20}$$⇒21C0​+21C1​+21C2​+.........+21C10​=220