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Question: If \({}^n{C_{r - 1}} = 10,{}^n{C_r} = 45\) and \({}^n{C_{r + 1}} = 120\) then r equals A.1 B.2 ...

If nCr−1=10,nCr=45{}^n{C_{r - 1}} = 10,{}^n{C_r} = 45 and nCr+1=120{}^n{C_{r + 1}} = 120 then r equals
A.1
B.2
C.3
D.4

Explanation

Solution

Firstly, using the formula nCrnCr−1=n−r+1r\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} and substituting the appropriate values find a linear equation in terms of n and r.
Then, by replacing r by r+1r + 1 in the equation nCrnCr−1=n−r+1r\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} and substituting the appropriate values find another linear equation in terms of n and r.
Finally, solve these linear equations to find the value of r.

Complete step-by-step answer:
Here, it is given that, nCr−1=10,nCr=45{}^n{C_{r - 1}} = 10,{}^n{C_r} = 45 and nCr+1=120{}^n{C_{r + 1}} = 120 .
Now, to solve we will use the formula nCrnCr−1=n−r+1r\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} . … (1)
So, substituting the values nCr−1=10,nCr=45{}^n{C_{r - 1}} = 10,{}^n{C_r} = 45 in the above equation.
⇒4510=n−r+1r ⇒92=n−r+1r ⇒9r=2(n−r+1) ⇒9r=2n−2r+2 ⇒2n−2r−9r+2=0  \Rightarrow \dfrac{{45}}{{10}} = \dfrac{{n - r + 1}}{r} \\\ \Rightarrow \dfrac{9}{2} = \dfrac{{n - r + 1}}{r} \\\ \Rightarrow 9r = 2\left( {n - r + 1} \right) \\\ \Rightarrow 9r = 2n - 2r + 2 \\\ \Rightarrow 2n - 2r - 9r + 2 = 0 \\\
⇒2n−11r+2=0\Rightarrow 2n - 11r + 2 = 0 … (2)
Now, in the equation (1), substituting r = r+1r + 1 , we get
nCr+1nCr+1−1=n−(r+1)+1r+1 ⇒nCr+1nCr=n−r−1+1r+1 ⇒nCr+1nCr=n−rr+1  \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_{r + 1 - 1}}}} = \dfrac{{n - \left( {r + 1} \right) + 1}}{{r + 1}} \\\ \Rightarrow \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} = \dfrac{{n - r - 1 + 1}}{{r + 1}} \\\ \Rightarrow \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} = \dfrac{{n - r}}{{r + 1}} \\\
Now, substituting the values of nCr=45{}^n{C_r} = 45 and nCr+1=120{}^n{C_{r + 1}} = 120 in the above equation we get
12045=n−rr+1 ⇒83=n−rr+1 ⇒8(r+1)=3(n−r) ⇒8r+8=3n−3r ⇒3n−3r−8r−8=0  \dfrac{{120}}{{45}} = \dfrac{{n - r}}{{r + 1}} \\\ \Rightarrow \dfrac{8}{3} = \dfrac{{n - r}}{{r + 1}} \\\ \Rightarrow 8\left( {r + 1} \right) = 3\left( {n - r} \right) \\\ \Rightarrow 8r + 8 = 3n - 3r \\\ \Rightarrow 3n - 3r - 8r - 8 = 0 \\\
⇒3n−11r−8=0\Rightarrow 3n - 11r - 8 = 0 … (3)
Now, we will solve equations (2) and (3) to get the value of r.
On multiplying equation (2) by 3 and equation (3) by -2, we get
6n−33r+6=0 \-6n+22r+16=0  6n - 33r + 6 = 0 \\\ \- 6n + 22r + 16 = 0 \\\
On solving the above pair of linear equations, we will get the required answer.
⇒−11r+22=0 ⇒−11(r−2)=0 ⇒r−2=0 ⇒r=2  \Rightarrow - 11r + 22 = 0 \\\ \Rightarrow - 11\left( {r - 2} \right) = 0 \\\ \Rightarrow r - 2 = 0 \\\ \Rightarrow r = 2 \\\
Thus, we get the values of r as 2.
So, option (B) is correct.

Note: Proof that nCrnCr−1=n−r+1r\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} .
We know that, nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} and nCr−1=n!(r−1)!(n−r+1)!{}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}
Now, taking ratio nCrnCr−1\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} , we get
nCrnCr−1=n!r!(n−r)!n!(r−1)!(n−r+1)! =(r−1)!(n−r+1)!r!(n−r)! =(r−1)!(n−r+1)(n−r)!r(r−1)!(n−r)! =n−r+1r  \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}}} \\\ = \dfrac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}{{r!\left( {n - r} \right)!}} \\\ = \dfrac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)\left( {n - r} \right)!}}{{r\left( {r - 1} \right)!\left( {n - r} \right)!}} \\\ = \dfrac{{n - r + 1}}{r} \\\