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Question: If \[{}^{n}{{C}_{r-1}}=36,{}^{n}{{C}_{r}}=84\ and\ {}^{n}{{C}_{r+1}}=126\] , then find the values of...

If nCr−1=36,nCr=84 and nCr+1=126{}^{n}{{C}_{r-1}}=36,{}^{n}{{C}_{r}}=84\ and\ {}^{n}{{C}_{r+1}}=126 , then find the values of n and r by evaluating the three equations.

Explanation

Solution

HINT: - The expansion of nCr{}^{n}{{C}_{r}} is =n!r!(n−r)!=\dfrac{n!}{r!\left( n-r \right)!} (Where n are r are the same numbers as in nCr{}^{n}{{C}_{r}} )

Complete step-by-step answer:
As mentioned in the question, the first equation can be written using the formula mentioned in the as
nCr−1=n!(r−1)!(n−(r−1))!=36     ...(a){}^{n}{{C}_{r-1}}=\dfrac{n!}{(r-1)!\left( n-(r-1) \right)!}=36\ \ \ \ \ ...\left( a \right)
The second equation can be written using the formula mentioned in the hint as
nCr=n!r!(n−r)!=84     ...(b){}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=84\ \ \ \ \ ...\left( b \right)
Similarly, the third equation can be written using the help of the formula mentioned in the hint as
nCr+1=n!(r+1)!(n−(r+1))!=126     ...(c){}^{n}{{C}_{r+1}}=\dfrac{n!}{(r+1)!\left( n-(r+1) \right)!}=126\ \ \ \ \ ...(c)

Now, using the equations above mentioned that are (a), (b) and (c),
On dividing (a) and (b), we get
n!(r−1)!(n−(r−1))!n!r!(n−r)!=3684\dfrac{\dfrac{n!}{(r-1)!\left( n-(r-1) \right)!}}{\dfrac{n!}{r!\left( n-r \right)!}}=\dfrac{36}{84}
On simplifying the above equation, we get
r(n−r+1)=37\dfrac{r}{(n-r+1)}=\dfrac{3}{7}
On cross multiplying, we get
10r=3n+3    ...(d)10r=3n+3\ \ \ \ ...(d)
Similarly, on dividing (b) and (c), we get
n!r!(n−r)!n!(r+1)!(n−(r+1))!=84126\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{(r+1)!\left( n-(r+1) \right)!}}=\dfrac{84}{126}
On simplifying the above equation, we get
(r+1)(n−r)=23\dfrac{\left( r+1 \right)}{(n-r)}=\dfrac{2}{3}
On cross multiplying, we get
5r=2n−3     (e)5r=2n-3\ \ \ \ \ (e)
Now, to get the value or to find out the value of n and r that has been used in these equations, we can perform elimination method to get the values.
So,
On multiplying equation (e) with 2, we get
10r=4n−6     ...(f)10r=4n-6\ \ \ \ \ ...(f)
Now, on equating equation (d) and (f), we get

& 3n+3=4n-6 \\\ & n=9 \\\ \end{aligned}$$ Now, we have got the value of n, so we can just put the value of n in any of the equations and then get the value of r. So, on putting the value of n in (f), we get $$\begin{aligned} & 10r=4\times 9-6 \\\ & 10r=36-6 \\\ & 10r=30 \\\ & r=3 \\\ \end{aligned}$$ Hence, the value of n = 9 and the value of r = 3. $$$$ NOTE:The students can make an error in evaluating the value of $${}^{n}{{C}_{r}}$$ and might confuse with $${}^{n}{{P}_{r}}$$ which has a similar expansion to $${}^{n}{{C}_{r}}$$ that is $$\dfrac{n!}{r!}$$ which will give a wrong answer when the question is done using it. .