Question

Given that \[\dfrac{\left( 2n \right)!}{7!\left( 2n-7 \right)!}:\dfrac{n!}{4!\left( n-4 \right)!}=24:1\] then find the value of n.

Answer 1

Hint: Now we are given the ratio of two expressions, first consider the numerator. Now we know that $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$, using this we will expand $\left( 2n \right)!$ such that we can cancel the denominator. Now from the even terms we will take 2 common. Now we will consider $\dfrac{n!}{4!\left( n-4 \right)!}$ . Again we will use $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$ to expand the numerator such that the denominator gets cancelled out. Now we will divide the obtained expressions and equate it to the given ratio to find n.

Complete step by step answer:
Now first let us consider $\dfrac{\left( 2n \right)!}{7!\left( 2n-7 \right)!}$
Now we know that $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$
Hence we can write $\left( 2n \right)!$ as
$\left( 2n \right)!=\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)\left( 2n-3 \right)\left( 2n-4 \right)\left( 2n-5 \right)\left( 2n-6 \right)\left( 2n-7 \right)!$
Hence using this we get
$\dfrac{\left( 2n \right)!}{7!\left( 2n-7 \right)!}=\dfrac{\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)\left( 2n-3 \right)\left( 2n-4 \right)\left( 2n-5 \right)\left( 2n-6 \right)\left( 2n-7 \right)!}{7!\left( 2n-7 \right)!}$
Now cancelling $\left( 2n-7 \right)!$ we get.
$=\dfrac{\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)\left( 2n-3 \right)\left( 2n-4 \right)\left( 2n-5 \right)\left( 2n-6 \right)}{7!}$
$=\dfrac{2\left( n \right)2\left( n-1 \right)2\left( n-2 \right)2\left( n-3 \right)\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7!}$
$=\dfrac{16n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7!}$
Let us call this equation as equation (1)
$\dfrac{\left( 2n \right)!}{7!\left( 2n-7 \right)!}=\dfrac{16n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7!}............................\left( 1 \right)$
Now consider $\dfrac{n!}{4!\left( n-4 \right)!}$
Again, in the equation if will write $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)!$ we get
$\dfrac{n!}{4!\left( n-4 \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)!}{4!\left( n-4 \right)!}$
$\dfrac{n!}{4!\left( n-4 \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{4!}............................\left( 2 \right)$
Now from dividing equation (1) by equation (2) we get.
$\dfrac{\dfrac{\left( 2n \right)!}{7!\left( 2n-7 \right)!}}{\dfrac{n!}{4!\left( n-4 \right)!}}=\dfrac{\dfrac{16n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7!}}{\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{4!}}$
$=\dfrac{\dfrac{16\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7\times 6\times 5\times 4!}}{\dfrac{1}{4!}}$
$=\dfrac{\dfrac{16\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7\times 6\times 5}}{1}$
Now this ratio is given to be 24 : 1
Hence we get \[\dfrac{16\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)}{7\times 6\times 5}=24\]
Now dividing the above equation by 8 we get.
\[\dfrac{2\left( 2n-1 \right)\left( 2n-5 \right)\left( 2n-3 \right)}{7\times 6\times 5}=3\]
Now multiplying the equation by 7 × 6 × 5 we get
$\begin{align}
  & 2\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)=7\times 6\times 5\times 3 \\
 & \Rightarrow 2\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)=630 \\
\end{align}$
Now again dividing the equation by 2 we get
$\left( 2n-1 \right)\left( 2n-3 \right)\left( 2n-5 \right)=315$
Now we know that $\left( 2n-1 \right),\left( 2n-3 \right)$ and $\left( 2n-5 \right)$ are nothing but odd consecutive numbers.
Hence we need to find three odd consecutive factors of 315
Now let us first check the prime factorization of 315.
$315=3\times 3\times 5\times 7$
Now by observation we can see that 5 × 7 × 9 = 315.
$\left( 2n-1 \right)$ is the smallest among $\left( 2n-1 \right),\left( 2n-3 \right)$ and $\left( 2n-5 \right)$
Hence we get $2n-1=5$
$\Rightarrow 2n=6$
Dividing the equation by 6 we get
$n=3$

Hence we have the value of n is 3.

Note: Note that in the expression the numerator is in the form of $\left( 2n \right)!$ here in factorial we cannot take 2 common as $\left( 2n \right)!\ne 2n!$ . Also in these questions when we get the final expression in n we need not form a cubic equation to solve. Just by observation we can find the value of n.