Question

What is \[C\left( 47,4 \right)+C\left( 51,3 \right)+C\left( 50,3 \right)+C\left( 49,3 \right)+C\left( 48,3 \right)+C\left( 47,3 \right)\] equal to?
A. $C\left( 47,4 \right)$
B. $C\left( 52,5 \right)$
C. $C\left( 52,4 \right)$
D. $C\left( 47,5 \right)$

Answer 1

Hint: C(n, r) is the notation used for ${}^{n}{{C}_{r}}$.
Calculate C(47, 4) + C(47, 3) using $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$ and then using the same identity add the obtained with C(49, 3) and using the same identity three more times obtain the required value of

Complete step-by-step answer:
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}\].

According to question, we have to find the value of \[C\left( 47,4 \right)+C\left( 51,3 \right)+C\left( 50,3 \right)+C\left( 49,3 \right)+C\left( 48,3 \right)+C\left( 47,3 \right)\]
C(n, r) is the notation used for ${}^{n}{{C}_{r}}$.
So, we have to find the value of \[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}\]
Rearranging the terms, we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+\left( {}^{47}{{C}_{3}}+{}^{47}{{C}_{4}} \right)+{}^{48}{{C}_{3}}\]
We know that $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$
Putting n = 47 and r = 4,
${}^{47}{{C}_{4}}+{}^{47}{{C}_{3}}={}^{48}{{C}_{4}}$
Replacing $''{}^{47}{{C}_{4}}+{}^{47}{{C}_{3}}''\ with\ ''{}^{48}{{C}_{4}}''$, we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{4}}+{}^{48}{{C}_{3}}\]
Again, we will apply $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$ for ${}^{48}{{C}_{3}}+{}^{48}{{C}_{4}}$.
Putting n = 48, r = 4 in the above formula, we will get,
${}^{48}{{C}_{3}}+{}^{48}{{C}_{4}}={}^{49}{{C}_{4}}$
On putting $''{}^{48}{{C}_{3}}+{}^{48}{{C}_{4}}={}^{49}{{C}_{4}}''$in above equation, we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{49}{{C}_{4}}\]
Again, we will apply the same formula for ${}^{49}{{C}_{3}}+{}^{49}{{C}_{4}}$.
Putting n = 49, r = 4 in the formula $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{50}{{C}_{4}}\]
Again, we will apply the same formula for ${}^{50}{{C}_{3}}+{}^{50}{{C}_{4}}$.
Putting n = 50, r = 4 in the formula $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$we will get,
\[{}^{50}{{C}_{4}}+{}^{50}{{C}_{3}}={}^{51}{{C}_{4}}\]
Now, replacing \[{}^{50}{{C}_{4}}+{}^{50}{{C}_{3}}\ with\ {}^{51}{{C}_{4}}\] in above equation, we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{51}{{C}_{3}}+{}^{51}{{C}_{4}}\]
Now, we will apply the same formula one more time for \[{}^{51}{{C}_{3}}+{}^{51}{{C}_{4}}\]. Putting n = 51, r = 4 in the formula $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$we will get,
\[{}^{51}{{C}_{4}}+{}^{51}{{C}_{3}}={}^{52}{{C}_{4}}\]
Now, replacing \[{}^{51}{{C}_{4}}+{}^{51}{{C}_{3}}\ with\ {}^{52}{{C}_{4}}\] in above equation, we will get,
\[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}={}^{52}{{C}_{4}}\]
Hence, the required value of \[{}^{47}{{C}_{4}}+{}^{51}{{C}_{3}}+{}^{50}{{C}_{3}}+{}^{49}{{C}_{3}}+{}^{48}{{C}_{3}}+{}^{47}{{C}_{3}}\ is{}^{52}{{C}_{4}}\] and we can write \[{}^{52}{{C}_{4}}=C\left( 52,4 \right)\].
Option (C) is the correct answer.

Note:Memorize the identity of binomial theorem used in the solution, $''{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}''$. We can prove this identity in the following ways:
$LHS={}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}$
We know ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
$\Rightarrow LHS=\dfrac{n!}{r!\left( n-r \right)!}+\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}$
Taking $n!$ common, we will get,
$\Rightarrow LHS=n!\left[ \dfrac{1}{\left( n-r \right)!r!}+\dfrac{1}{\left( n-r+1 \right)!\left( r-1 \right)!} \right]$
Taking $\dfrac{1}{\left( r-1 \right)!}$ common, we get,
$\Rightarrow LHS=\dfrac{n!}{\left( r-1 \right)!}\left[ \dfrac{1}{\left( n-r \right)!r!}+\dfrac{1}{\left( n-r+1 \right)!} \right]$
We know, $n!=n\left( n-1 \right)!,\ so\ \left( n-r+1 \right)!=\left( n-r+1 \right)\left( n-r \right)!$
$\Rightarrow LHS=\dfrac{n!}{\left( r-1 \right)!}\left[ \dfrac{1}{r\left( n-r \right)!}+\dfrac{1}{\left( n-r+1 \right)\left( n-r \right)!} \right]$
Taking $\dfrac{1}{\left( n-r \right)!}$ common, we will get,
$\Rightarrow LHS=\dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{1}{r}+\dfrac{1}{n-r+1} \right]$
Taking LCM and adding, we will get,
$\begin{align}
  & \Rightarrow LHS=\dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{n-{r}+1+{r}}{r\left( n-r+1 \right)} \right] \\
 & \Rightarrow LHS=\dfrac{n!}{\left( r-1 \right)!\left( n-r \right)!}\left[ \dfrac{n+1}{r\left( n-r+1 \right)} \right] \\
 & \Rightarrow LHS=\dfrac{\left( n+1 \right)n!}{\left( r-1 \right)!\left( n-r+1 \right)\left( n-r \right)!} \\
\end{align}$
We know,
 $\begin{align}
  & \left( n+1 \right)\times n!=\left( n+1 \right)! \\
 & \Rightarrow r\left( r-1 \right)!=r!\ and \\
 & \left( n-r+1 \right)\left( n-r \right)!=\left( n-r+1 \right)! \\
\end{align}$
Thus, we will get,
\[\begin{align}
  & LHS=\dfrac{\left( n+1 \right)!}{r!\left[ \left( n+1 \right)-r \right]!} \\
 & \Rightarrow LHS={}^{n+1}{{C}_{r}} \\
 & \Rightarrow LHS=RHS \\
\end{align}\]
Thus, we have proved that ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$.