Question

It is given that if ${}^{2n}{{C}_{3}}:{}^{n}{{C}_{2}}=44:3$ then for which value of r the value of ${}^{n}{{C}_{r}}$ will be 15?
(a) $r=3$
(b) $r=4$
(c) $r=6$
(d) $r=5$

Answer 1

We have given the equation ${}^{2n}{{C}_{3}}:{}^{n}{{C}_{2}}=44:3$ so in this equation first of all we will expand the left hand side by using the relation of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ then after expanding the left hand side of the equation solve the equation by simple arrangement and basic algebra to get the value of n. Now, we have to find the value of r when ${}^{n}{{C}_{r}}=15$ so expand the left hand side of this equation in the way that we have shown above and substitute the value of n and solve. Solving will give you the value of “r”.

Complete step by step answer:
We have given the combinatorial relation as follows:
${}^{2n}{{C}_{3}}:{}^{n}{{C}_{2}}=44:3$……….. Eq. (1)
We know that,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above relation we can write ${}^{2n}{{C}_{3}}\And {}^{n}{{C}_{2}}$ as follows:
$\begin{aligned} & {}^{2n}{{C}_{3}}=\dfrac{2n!}{3!\left( 2n-3 \right)!} \\\ & \Rightarrow {}^{2n}{{C}_{3}}=\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)\left( 2n-3 \right)!}{3!\left( 2n-3 \right)!} \\\ \end{aligned}$
In the above equation, $\left( 2n-3 \right)!$ will be cancelled out from the numerator and denominator of the right hand side of the above equation so the remaining expression will look like:
${}^{2n}{{C}_{3}}=\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{3!}$
We can write 3! as 3.2.1 in the above equation,
$\begin{aligned} & {}^{2n}{{C}_{3}}=\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{3.2.1} \\\ & \Rightarrow {}^{2n}{{C}_{3}}=\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{6} \\\ \end{aligned}$
${}^{n}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}$
Now, substituting the values of ${}^{n}{{C}_{2}}\And {}^{2n}{{C}_{3}}$ in eq. (1) we get,
${}^{2n}{{C}_{3}}:{}^{n}{{C}_{2}}=44:3$
$\Rightarrow \dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{6}:\dfrac{n\left( n-1 \right)}{2}=44:3$
Eliminating the ratio sign by putting division sign in place of ratio in the above equation we get,
$\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{6\left( n\left( n-1 \right) \right)}\times 2=\dfrac{44}{3}$
2 divide 6 by 3 times so the above equation will look like:
$\begin{aligned} & \dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{3\left( n\left( n-1 \right) \right)}=\dfrac{44}{3} \\\ & =\dfrac{2n\left( 2n-1 \right)\left( 2n-2 \right)}{\left( n\left( n-1 \right) \right)}=\dfrac{44}{1} \\\ \end{aligned}$
“n” will be cancelled out from the numerator and the denominator of the above equation.
$\dfrac{2\left( 2n-1 \right)\left( 2n-2 \right)}{\left( n-1 \right)}=\dfrac{44}{1}$
Taking 2 as common from $\left( 2n-2 \right)$ we get,
$\begin{aligned} & \dfrac{2\left( 2 \right)\left( 2n-1 \right)\left( n-1 \right)}{\left( n-1 \right)}=\dfrac{44}{1} \\\ & \Rightarrow \dfrac{4\left( 2n-1 \right)}{1}=44 \\\ \end{aligned}$
Dividing 4 on both the sides of the above equation we get,
$\begin{aligned} & \left( 2n-1 \right)=\dfrac{44}{4} \\\ & \Rightarrow \left( 2n-1 \right)=11 \\\ \end{aligned}$
Adding 1 on both the sides of the above equation we get,
$2n=12$
Dividing 2 on both the sides we get,
$\begin{aligned} & n=\dfrac{12}{2} \\\ & \Rightarrow n=6 \\\ \end{aligned}$
Hence, we have got the value of n as 6.
Now, substituting this value of n in ${}^{n}{{C}_{r}}$ and it is given that ${}^{n}{{C}_{r}}=15$ then equating ${}^{n}{{C}_{r}}$ to 15 we get,
${}^{6}{{C}_{r}}=15$
Expanding the left hand side of the above equation we get,
$\dfrac{6!}{r!\left( 6-r \right)!}=15$
Rearranging the above equation we get,
$\begin{aligned} & \dfrac{6.5.4.3.2.1}{15}=r!\left( 6-r \right)! \\\ & \Rightarrow \dfrac{48\left( 15 \right)}{15}=r!\left( 6-r \right)! \\\ \end{aligned}$
15 will be cancelled out from the numerator and denominator we get,
$48=r!\left( 6-r \right)!$…….Eq. (2)
Substituting $r=1$ in the above equation we get,
$\begin{aligned} & 48=1!\left( 6-1 \right)! \\\ & \Rightarrow 48=5! \\\ & \Rightarrow 48=120 \\\ \end{aligned}$
L.H.S is not equal to R.H.S, r = 1 is not satisfying.
Now, substituting $r=2$ in eq. (2) we get,
$\begin{aligned} & 48=2!\left( 6-2 \right)! \\\ & \Rightarrow 48=2\left( 4 \right)! \\\ & \Rightarrow 48=2\left( 24 \right) \\\ & \Rightarrow 48=48 \\\ \end{aligned}$
Hence, the value of $r=2$ is satisfying the eq. (2).
From the above solution, we have got the value of n as 6 and r as 2.
But on comparing the value of r as 2 from the options given above, none of the options are correct.
We know that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ so ${}^{6}{{C}_{2}}={}^{6}{{C}_{4}}$ from this we have got the value of r as 4.

So, the correct answer is “Option b”.

Note: The other way of solving the above problem is that after finding the value of n instead of substituting the value of r as 1 or 2 we can substitute the values of r given in the options.
For instance, checking the option (a) in ${}^{6}{{C}_{r}}$ we get,
Option (a) is $r=3$ so substituting the value of r as 3 in ${}^{6}{{C}_{r}}$ we get,
$\begin{aligned} & {}^{6}{{C}_{3}} \\\ & =\dfrac{6!}{3!3!} \\\ & =\dfrac{6.5.4.3!}{3!3!} \\\ & =\dfrac{6.5.4}{6} \\\ & =20 \\\ \end{aligned}$
It is given that ${}^{6}{{C}_{r}}=15$ but on substituting the value of r as 3 we were getting the answer as 20 which is not equal to 15 so $r=3$ is not the correct option.
Similarly, you can check the other options too.