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

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

& \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{aligned}$$ ……………………………………(5) The above equation needs to be more simplified. We can write $$\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{aligned} & \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{aligned}$$ $$\Rightarrow 7r=3n-4$$ …………………………………(8) Similarly, applying the formula shown in equation (4) and on simplifying equation (3), we get $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \Rightarrow 63r-63r=28n-98-\left( 27n-36 \right) \\\ & \Rightarrow 0=28n-98-27n+36 \\\ & \Rightarrow 0=n-62 \\\ \end{aligned}$$ $$\Rightarrow 62=n$$ …………………………………………(15) On putting $$n=62$$ in equation (12), we get $$\begin{aligned} & \Rightarrow 9r=4\times 62-14 \\\ & \Rightarrow 9r=234 \\\ & \Rightarrow r=\dfrac{234}{9} \\\ \end{aligned}$$ $$\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{aligned} & =2\times 62+3\times 62 \\\ & =124+186 \\\ & =310 \\\ \end{aligned}$$ **Therefore, the value of $$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.