Question

If we have the binomial coefficient as \[^{n}{{C}_{r}}{{:}^{n}}{{C}_{r+1}}{{:}^{n}}{{C}_{r+2}}=3:4:5\] , then the value of \[2n+3r\] is
(A) 238
(B) 220
(C) 203
(D) 240

Answer 1

Hint: First of all, split the ratio as \[^{n}{{C}_{r}}{{:}^{n}}{{C}_{r+1}}=3:4\] and \[^{n}{{C}_{r+1}}{{:}^{n}}{{C}_{r+2}}=4:5\] . Now, use the formula \[^{n}{{C}_{r}}\] , \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and modify the ratios. We can write \[\left( r+1 \right)!\] as a product of \[r!\] and \[\left( r+1 \right)\] . Similarly, \[\left( n-r \right)!\] can be written as a product of \[\left( n-r-1 \right)!\] and \[\left( n-r \right)\]. Now, solve it further and get the value of \[n\] and \[r\] . Using the value of \[n\] and \[r\] calculate the value of \[2n+3r\].

Complete step-by-step solution
According to the question, we are given that
\[^{n}{{C}_{r}}{{:}^{n}}{{C}_{r+1}}{{:}^{n}}{{C}_{r+2}}=3:4:5\] ……………………………….(1)
Let us split the above ratio.
On splitting, we get
\[^{n}{{C}_{r}}{{:}^{n}}{{C}_{r+1}}=3:4\] ……………………………………(2)
 \[^{n}{{C}_{r+1}}{{:}^{n}}{{C}_{r+2}}=4:5\] ………………………………………….(3)
We know the formula for \[^{n}{{C}_{r}}\] , \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] …………………………………….(4)
Now, applying the formula shown in equation (4) and on simplifying equation (2), we get
\[\begin{align}
  & \Rightarrow \dfrac{^{n}{{C}_{r}}}{^{n}{{C}_{r+1}}}=\dfrac{3}{4} \\
 & \Rightarrow \dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}}=\dfrac{3}{4} \\
\end{align}\]
 ……………………………………(5)
The above equation needs to be more simplified.
We can write \[\left( r+1 \right)!\] as product of \[r!\] and \[\left( r+1 \right)\] ………………………………..(6)
Similarly, \[\left( n-r \right)!\] can be written as product of \[\left( n-r-1 \right)!\] and \[\left( n-r \right)\] ……………………………………(7)
Now, from equation (5), equation (6), and equation (7), we get
\[\begin{align}
  & \Rightarrow \dfrac{r!\left( r+1 \right)\left( n-r-1 \right)!}{r!\left( n-r-1 \right)!\left( n-r \right)}=\dfrac{3}{4} \\
 & \Rightarrow \dfrac{\left( r+1 \right)}{\left( n-r \right)}=\dfrac{3}{4} \\
 & \Rightarrow 4r+4=3n-3r \\
 & \Rightarrow 4r+3r=3n-4 \\
\end{align}\]
\[\Rightarrow 7r=3n-4\] …………………………………(8)
Similarly, applying the formula shown in equation (4) and on simplifying equation (3), we get
 \[\begin{align}
  & \Rightarrow \dfrac{^{n}{{C}_{r+1}}}{^{n}{{C}_{r+2}}}=\dfrac{4}{5} \\
 & \Rightarrow \dfrac{\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}}{\dfrac{n!}{\left( r+2 \right)!\left( n-r-2 \right)!}}=\dfrac{4}{5} \\
\end{align}\]
\[\Rightarrow \dfrac{\left( r+2 \right)!\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-1 \right)!}=\dfrac{4}{5}\] ……………………………………(9)
The above equation needs to be more simplified.
We can write \[\left( r+2 \right)!\] as product of \[\left( r+1 \right)!\] and \[\left( r+2 \right)\] ………………………………..(10)
Similarly, \[\left( n-r-1 \right)!\] can be written as product of \[\left( n-r-2 \right)!\] and \[\left( n-r-1 \right)\] ……………………………………(11)
Now, from equation (9), equation (10), and equation (11), we get
\[\begin{align}
  & \Rightarrow \dfrac{\left( r+1 \right)!\left( r+2 \right)\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-2 \right)!\left( n-r-1 \right)}=\dfrac{4}{5} \\
 & \Rightarrow \dfrac{\left( r+2 \right)}{\left( n-r-1 \right)}=\dfrac{4}{5} \\
 & \Rightarrow 5r+10=4n-4r-4 \\
 & \Rightarrow 5r+4r=4n-4-10 \\
\end{align}\]
\[\Rightarrow 9r=4n-14\] …………………………………(12)
On multiplying equation (8) by 9, we get
\[\Rightarrow 7r\times 9=\left( 3n-4 \right)\times 9\]
\[\Rightarrow 63r=27n-36\]…………………………………………………..(13)
On multiplying equation (12) by 7, we get
\[\Rightarrow 9r\times 7=\left( 4n-14 \right)\times 7\]
\[\Rightarrow 63r=28n-98\] ………………………………………………….(14)
Similarly, on subtracting equation (13) from equation (14), we get
\[\begin{align}
  & \Rightarrow 63r-63r=28n-98-\left( 27n-36 \right) \\
 & \Rightarrow 0=28n-98-27n+36 \\
 & \Rightarrow 0=n-62 \\
\end{align}\]
\[\Rightarrow 62=n\] …………………………………………(15)
On putting \[n=62\] in equation (12), we get
\[\begin{align}
  & \Rightarrow 9r=4\times 62-14 \\
 & \Rightarrow 9r=234 \\
 & \Rightarrow r=\dfrac{234}{9} \\
\end{align}\]
\[\Rightarrow r=26\] ………………………………………(16)
We are asked to find the value of \[2n+3r\] ……………………………….(17)
Now, from equation (15), equation (16), and equation (17), we get
\[\begin{align}
  & =2\times 62+3\times 62 \\
 & =124+186 \\
 & =310 \\
\end{align}\]
Therefore, the value of \[2n+3r\] is 310.

Note: For this type of question, where we have some expression in terms of \[^{n}{{C}_{r}}\] . The best way to approach this type of question is to follow the basic formula for \[^{n}{{C}_{r}}\] , \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] . Use this formula and modify the expression to simplify it into simpler form.