Question

The sum ${}^{10}{{C}_{3}}+{}^{11}{{C}_{3}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}$is equal to:
(a) ${}^{21}{{C}_{4}}$
(b) ${}^{21}{{C}_{4}}-{}^{10}{{C}_{4}}$
(c) ${}^{21}{{C}_{4}}-{}^{11}{{C}_{4}}$
(d) ${}^{21}{{C}_{17}}$

Answer 1

Hint:In the given sum of series, add and subtract${}^{10}{{C}_{4}}$ then you will find in the series two terms ${}^{10}{{C}_{3}}+{}^{10}{{C}_{4}}$ is resolved to ${}^{11}{{C}_{4}}$ by using the relation ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$, then club this term with next term ${}^{11}{{C}_{3}}$ ,Again use above relation you will get ${}^{12}{{C}_{4}}$. Similarly, do this till ${}^{20}{{C}_{3}}+{}^{20}{{C}_{4}}$.

Complete step-by-step answer:
The summation of the series that we are asked to find is:
  ${}^{10}{{C}_{3}}+{}^{11}{{C}_{3}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}$
Adding and subtracting ${}^{10}{{C}_{4}}$ in the above summation series we get,
  ${}^{10}{{C}_{3}}+{}^{11}{{C}_{3}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}+{}^{10}{{C}_{4}}-{}^{10}{{C}_{4}}$
We are going to write ${}^{10}{{C}_{4}}$adjacent to${}^{10}{{C}_{3}}$
${}^{10}{{C}_{3}}+{}^{10}{{C}_{4}}+{}^{11}{{C}_{3}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}-{}^{10}{{C}_{4}}$
In the above summation, you will see that ${}^{10}{{C}_{3}}+{}^{10}{{C}_{4}}$ are satisfying this relation ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ where n = 10 and r = 4. So, using this relation we can write the above summation as:
${}^{11}{{C}_{4}}+{}^{11}{{C}_{3}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}-{}^{10}{{C}_{4}}$
Similarly, ${}^{11}{{C}_{4}}+{}^{11}{{C}_{3}}$ is also satisfying the relation ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ then the above summation will be written as follows:
${}^{12}{{C}_{4}}+{}^{12}{{C}_{3}}+.........+{}^{20}{{C}_{3}}-{}^{10}{{C}_{4}}$
Similarly, you can use this relation till ${}^{20}{{C}_{3}}$ and the summation till ${}^{20}{{C}_{3}}$ is resolved to ${}^{21}{{C}_{4}}$ i.e ${}^{20}{{C}_{3}}+{}^{20}{{C}_{4}}$=${}^{21}{{C}_{4}}$
Hence, the above summation will be written as:
${}^{21}{{C}_{4}}-{}^{10}{{C}_{4}}$
So, the summation of the given series is equal to ${}^{21}{{C}_{4}}-{}^{10}{{C}_{4}}$.
Hence, the correct option is (b).

Note: In the above solution, you might be thinking that how do we know when we have to subtract${}^{10}{{C}_{4}}$from the summation. As you will see in the series, the number in the superscript of C is increasing consecutively and the subscript remains same so if we add and subtract${}^{10}{{C}_{4}}$then we can resolve the summation in the form of this relation${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$.So students should remember this relation for solving these types of questions.