Question

If ${}^n{C_{r - 1}} = 36,{}^n{C_r} = 84,{}^n{C_{r + 1}} = 126$, find the value of ${}^r{C_2}$.

Answer 1

We know that ${}^n{C_r} = \dfrac{{n!}}{{n - r!r!}}$ and $n! = n(n - 1)(n - 2)...........3.2.1$ and we need to find the value of ${}^r{C_2} = \dfrac{{r!}}{{r - 2!2!}}$ so we need to find the value of $r$.

Complete step by step solution:
Here we are given the value of ${}^n{C_{r - 1}} = 36,{}^n{C_r} = 84,{}^n{C_{r + 1}} = 126$ and now we should know what is ${}^n{C_r}$ so here ${}^n{C_r}$is the combination of different things from n things or in other words we can say that ${}^n{C_r}$ is the number of different unordered combinations of $r$ objects from the set of $n$ objects. We know the formula of ${}^n{C_r} = \dfrac{{n!}}{{n - r!r!}}$
Here $n! = n(n - 1)(n - 2)...........3.2.1$ denotes the factorial of the number$n$ and we should know that $0! = 1,1! = 1,2! = 2.1 = 2......{\text{ and so on}}$
Here n must be the positive integer.
So now we are given that
$\Rightarrow {}^n{C_{r - 1}} = 36$ that is $\dfrac{{n!}}{{n - r + 1!r - 1!}} = 36$
Or we can write it as
$\Rightarrow \dfrac{{n!}}{{(n + 1 - r)!r - 1!}} = 36$ $- - - - (1)$
And we know that ${}^n{C_r} = 84$
$\Rightarrow \dfrac{{n!}}{{n - r!r!}} = 84$ $- - - - (2)$
Now dividing (1) and (2) we get that
$\Rightarrow \dfrac{{(n - r)!r(r - 1)!}}{{(n + 1 - r)(n - r)!(r - 1)!}} = \dfrac{3}{7}$
So we get that
$\Rightarrow \dfrac{r}{{(n + 1 - r)}} = \dfrac{3}{7}$
$\Rightarrow 7r = 3n + 3 - 3r$
$\Rightarrow 10r = 3n + 3$ $- - - - - (3)$
Also we are given that ${}^n{C_{r + 1}} = 126$
So we can write that
$\Rightarrow \dfrac{{n!}}{{n - r - 1!r + 1!}} = 84$ $- - - - - (4)$
Now dividing (2) and (3) we get that
$\Rightarrow \dfrac{{(n - r - 1)!(r + 1)(r)!}}{{(n - r)(n - r - 1)!(r)!}} = \dfrac{2}{3} \\\ \Rightarrow \dfrac{{(r + 1)}}{{(n - r)}} = \dfrac{2}{3} \\\ \Rightarrow 3r + 3 = 2n - 2r \\\$
$5r = 2n - 3 - - - - - - (5)$
Now solving (3) and (5)
Now multiplying the equation (5) by 2 and then subtracting it from equation (3) we get
$\Rightarrow 4n - 6 - 3n - 3 = 0 \\\ \Rightarrow n = 9 \\\$
Now putting it in (3) we get that
$\Rightarrow 10r = 3(9) + 3 \\\ \Rightarrow r = 3 \\\$

So we can write ${}^r{C_2} = \dfrac{{r!}}{{r - 2!2!}}$
${}^3{C_2} = \dfrac{{3!}}{{3 - 2!2!}} = \dfrac{6}{2} = 3$

Note:
We should know that for any $n \geqslant 1$ ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ........ + {}^n{C_n} = {2^n}$ It is valid for $n \geqslant 1$. Also, combinations are used to choose the elements. ${}^n{C_r}$ means, number of ways to choose r items from n.