Question

Find the value of $^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+…+^{5}C_{4}$
(a) $^{11}C_{5}$
(b) $^{11}C_{4}$
(c ) $^{11}C_{7}$
(d) $^{11}C_{5}-1$

Answer 1

Hint: In these types of question we need to use the property of combination which is $^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}$

Complete step-by-step answer:
Given, the series $^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+…+^{5}C_{4}$
Add $^{5}C_{5}$ and subtract 1 from the series to keep the series same
=$^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+^{7}C_{4}+^{6}C_{4}+^{5}C_{4}+^{5}C_{5}-1$
Use the property of combination,
=$^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+^{7}C_{4}+^{6}C_{4}+^{6}C_{5}-1$
Again take the fifth and sixth term together and use the property of combination,
=$^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+^{7}C_{4}+^{7}C_{5}-1$

Similarly take the fourth and fifth term together and use the property of combination,
=$^{10}C_{4}+^{9}C_{4}+^{8}C_{4}+^{8}C_{5}-1$

Again take the third and fourth term together and use the property of combination,
=$^{10}C_{4}+^{9}C_{4}+^{9}C_{5}-1$

Take the second and third term together and use the property of combination,
=$^{10}C_{4}+^{10}C_{5}-1$

Finally take the first 2 terms to reduce the series,
=$^{11}C_{5}-1$


Note: In such types of questions we need to add some value as a solution requirement and subtract 1 in such a way so that the series remains the same and we get to use the property of combination.