Question

If ${}^{2n}{C_3}:{}^n{C_2} = 44:3$ then $n$=
$A) 6$
$B) 7$
$C) 8$
$D) 9$

Answer 1

Hint: Here we have to find the value of \[n\].
It is given the combination to solve it using the combination formula and we split some terms and finally we get the required answer.

Formula used: That is ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ firstly then simplifying the equation and get the required result.

Complete step-by-step solution:
It is given that the question stated as $n \in \mathbb{N}$ such that ${}^{2n}{C_3}:{}^n{C_2} = 44:3$
Here we have to find out the value for $n$
Using the formula of${}^n{C_r}$, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
We can write the given ratio in an expression as:
$ \Rightarrow \dfrac{{2n!}}{{3!(2n - 3)!}}:\dfrac{{n!}}{{2!(n - 2)!}} = 44:3$
Or we can write the term on equivalently as a fraction:
We just divided the one fraction to the other one
$ \Rightarrow \dfrac{{\dfrac{{2n!}}{{3!(2n - 3)!}}}}{{\dfrac{{n!}}{{2!(n - 2)!}}}} = \dfrac{{44}}{3}$
Now we just reversed both the fractions
$ \Rightarrow \dfrac{{2n!}}{{3!(2n - 3)!}} \times \dfrac{{2!(n - 2)!}}{{n!}} = \dfrac{{44}}{3}$
And if we expand the above expression as follows:
$ \Rightarrow \dfrac{{(2n)(2n - 1)(2n - 2)(2n - 3)......1}}{{3!(2n - 3)!}} \times \dfrac{{2!(n - 2)!}}{{n(n - 1)(n - 2)....1}} = \dfrac{{44}}{3}$
Cancelling the same terms from the numerators and denominators we get:
$ \Rightarrow \dfrac{{(2n)(2n - 1)2(n - 1)}}{{3!}} \times \dfrac{{2!}}{{n(n - 1)}} = \dfrac{{44}}{3}$
Expanding the factorial term and we get
\[ \Rightarrow \dfrac{{2(n)(2n - 1)2(n - 1)}}{6} \times \dfrac{2}{{n(n - 1)}} = \dfrac{{44}}{3}\]
Again we cancel the same terms from the numerators and denominators, we get
$ \Rightarrow \dfrac{{4(2n - 1)}}{3} = \dfrac{{44}}{3}$
We can cancel the multiple of terms on the opposite sides like $44by4$ and $3by3$
Hence we found an equation
$ \Rightarrow 2n - 1 = 11$
On adding $ - 1$ on both side, we can have
\[ \Rightarrow 2n = 11 + 1\]
By applying adding operation in the right hand side
$ \Rightarrow 2n = 12$
On dividing $2$ on both side we get,
$ \Rightarrow n = \dfrac{{12}}{2}$
We get,
$\therefore n = 6$

Hence the correct option is $(A)$ that is $6$

Note: In this problem we have to verify the answer whether we found the value is correct or not
Verifying the answer:
It is given that ${}^{2n}{C_3}:{}^n{C_2} = 44:3$
$\dfrac{{4(2n - 1)}}{3} = \dfrac{{44}}{3}$
We just put the value of \[n = 6\] we get
$\dfrac{{4(2 \times 6 - 1)}}{3} = \dfrac{{44}}{3}$
On multiply the terms we get,
$\dfrac{{4(12 - 1)}}{3} = \dfrac{{44}}{3}$
On subtracting the terms,
$\dfrac{{4(11)}}{3} = \dfrac{{44}}{3}$
Let us multiply the numerator we get
$\dfrac{{44}}{3} = \dfrac{{44}}{3}$
Hence the left hand side is equal to the right hand side.