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) ${}^{21}{{C}_{17}}$

Question

Vedantu Content Team · Accepted Answer

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.

The sum 10C3+11C3+12C3+.........+20C3is equal to:(a) 21C4 (b) 21C4−10C4(c) 21C4−11C4(d) 21C17

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) $^{21} C_{17}$