Question

The value of $\sum\limits_{r=0}^{40}{r.{}^{40}{{C}_{r}}{}^{30}{{C}_{r}}}$ is:
A. $40.{}^{69}{{C}_{29}}$
B. $40.{}^{70}{{C}_{30}}$
C. ${}^{69}{{C}_{29}}$
D. ${}^{70}{{C}_{30}}$

Answer 1

To find the value of $\sum\limits_{r=0}^{40}{r.{}^{40}{{C}_{r}}{}^{30}{{C}_{r}}}$ , first consider the terms other than the summation. Now, expand ${}^{40}{{C}_{r}}$ using ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ and cancel the common terms. Then use the properties of combination on ${}^{30}{{C}_{r}}$. We will be using the formula ${}^{n}{{C}_{n-r}}={}^{n}{{C}_{r}}$ . Then after substitution, we will be taking the summation of the result. Use the properties of combination, i.e $\sum{{}^{n}{{C}_{r}}.{}^{m}{{C}_{k}}}={}^{m+n}{{C}_{r+k}}$ on the resulted equation to yield the final answer.

Complete step-by-step answer:
We need to find the value of $\sum\limits_{r=0}^{40}{r.{}^{40}{{C}_{r}}{}^{30}{{C}_{r}}}$ . For this, let us assume ${{T}_{r}}=r.{}^{40}{{C}_{r}}{}^{30}{{C}_{r}}$ .
Now let’s expand ${}^{40}{{C}_{r}}$ .
Since, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$, we get
$\Rightarrow {{T}_{r}}=r.\dfrac{40!}{r!(40-r)!}.{}^{30}{{C}_{r}}$
Expanding $r!$ in the denominator, the above equation becomes,
${{T}_{r}}=r.\dfrac{40!}{r(r-1)!(40-r)!}.{}^{30}{{C}_{r}}$
Now let us cancel $r$ from the numerator and denominator. Then, the above equation will become
$\Rightarrow {{T}_{r}}=\dfrac{40!}{(r-1)!(40-r)!}.{}^{30}{{C}_{r}}$
Now expand $40!$ .
$\Rightarrow {{T}_{r}}=40.\dfrac{39!}{(r-1)!(40-r)!}.{}^{30}{{C}_{r}}$
Now, $\dfrac{39!}{(r-1)!(40-r)!}$ can be written as ${}^{39}{{C}_{r-1}}.$ .
Therefore, the above equation becomes,
${{T}_{r}}=40.{}^{39}{{C}_{r-1}}.{}^{30}{{C}_{r}}...(i)$
We know that ${}^{n}{{C}_{n-r}}=\dfrac{n!}{(n-r)!(n-n+r)!}=\dfrac{n!}{(n-r)!r!}={}^{n}{{C}_{r}}$
Therefore, ${}^{30}{{C}_{r}}$ can be written as:
${}^{30}{{C}_{r}}={}^{30}{{C}_{30-r}}$
Now, by substituting this in equation $(i)$, we will get
${{T}_{r}}=40.{}^{39}{{C}_{r-1}}.{}^{30}{{C}_{30-r}}$
Now let’s evaluate $\sum\limits_{r=0}^{40}{{{T}_{r}}}$.
$\Rightarrow \sum\limits_{r=0}^{40}{40.{}^{39}{{C}_{r-1}}.{}^{30}{{C}_{30-r}}}$
Now take $40$ outside as it is a constant. Then we will get
$40\sum\limits_{r=0}^{40}{{}^{39}{{C}_{r-1}}.{}^{30}{{C}_{30-r}}}...(ii)$
We know that $\sum{{}^{n}{{C}_{r}}.{}^{m}{{C}_{k}}}={}^{m+n}{{C}_{r+k}}$
Therefore, equation $(ii)$ becomes
$40.{}^{39+30}{{C}_{r-1+30-r}}$
By adding, we get
$\Rightarrow 40.{}^{69}{{C}_{29}}$

So, the correct answer is “Option A”.

Note: The backbone of this solution is the properties of combination and you must be thorough with it. Do not cancel the common terms in ${}^{40}{{C}_{r}}$ and ${}^{30}{{C}_{r}}$ . In these types of problems, try to solve without taking the summation. Do not apply the summation at the beginning. All you have to do is to simplify the equation and then apply the properties. Be careful to use the equation of combination as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ instead of permutation.