Question

The value of $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{5}{{C}_{4}} $ is:
(a) $ {}^{11}{{C}_{5}} $
(b) $ {}^{11}{{C}_{4}} $
(c) $ {}^{11}{{C}_{7}} $
(d) $ {}^{11}{{C}_{5}}-1 $

Answer 1

Hint: To solve the above summation, we are going to use the relation which is $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ . To apply this relation we have to add or subtract the summation series by 1 then we can write 1 as $ {}^{5}{{C}_{5}} $ . As you can see that $ {}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} $ is in the form of the relation that we have described above in n and r. Now, using this relation we can write $ {}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} $ as $ {}^{6}{{C}_{5}} $ . Similarly, we can club $ {}^{6}{{C}_{5}} $ with $ {}^{6}{{C}_{4}} $ and use the relation $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ . Using this relation in a similar way to the other members of the summation series will give the required answer.

Complete step-by-step answer:
We are asked to find the value of:
 $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{6}{{C}_{4}}+{}^{5}{{C}_{4}} $
To evaluate the above series, we are going to use the below relation:
 $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ …………. Eq. (1)
To use the above relation we want at least two terms in the summation series having subscript difference of 1 for that to happen we are going to add or subtract 1 in the summation series.
 $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{6}{{C}_{4}}+{}^{5}{{C}_{4}}+1-1 $
To make the terms in the above series in the form of eq. (1) we will write 1 as $ {}^{5}{{C}_{5}} $ in the above series.
 $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{6}{{C}_{4}}+{}^{5}{{C}_{4}}+{}^{5}{{C}_{5}}-1 $
Now, in the above series if you look carefully then you will find that $ {}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} $ is the form of eq. (1) so applying eq. (1) in $ {}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} $ we get,
 $ {}^{5}{{C}_{4}}+{}^{5}{{C}_{5}}={}^{6}{{C}_{5}} $
Substituting the above result in the summation series we get,
 $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{6}{{C}_{4}}+{}^{6}{{C}_{5}}-1 $
Similarly we can use the relation $ {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} $ in the above series in $ {}^{6}{{C}_{4}}+{}^{6}{{C}_{5}} $ .
 $ {}^{10}{{C}_{4}}+{}^{9}{{C}_{4}}+.......+{}^{7}{{C}_{5}}-1 $
After applying eq. (1) to the other combinatorial terms in the same as above we are left with:
 $ \begin{align}
  & {}^{10}{{C}_{4}}+{}^{10}{{C}_{5}}-1 \\
 & ={}^{11}{{C}_{5}}-1 \\
\end{align} $
From the above, the summation of series that we have got is $ {}^{11}{{C}_{5}}-1 $ .
Hence, the correct option is (d).

Note: You might have thought instead of solving the question in a way that we have shown above why we expand the individual terms in the summation series using the expansion of $ {}^{n}{{C}_{r}} $ which is:
 $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ ………..Eq. (2)
Expanding individual terms meaning using the above relation find the value of $ {}^{10}{{C}_{4}} $ and then add its result with $ {}^{9}{{C}_{4}} $ . Similarly, do this expansion and addition till $ {}^{5}{{C}_{4}} $ and note down the result and then check the options by expanding the options using eq. (2) and then compare the result. This way of solving this problem is a very lengthy process so it’s better to solve the problem in the manner that we have shown in the solution part.