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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\...

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 rr equals
A. 11
B. 22
C. 33
D. 44

Explanation

Solution

In this problem, we will use the properties of combination. We will use the property nCrnCr−1=n−r+1r\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} of combination. After using this property, we will get two equations in two variables. We will solve these two equations by a simple elimination method.

Complete step by step solution
In this problem, it is given that
nCr−1=10⋯⋯(1){}^n{C_{r - 1}} = 10 \cdots \cdots \left( 1 \right)
nCr=45⋯⋯(2){}^n{C_r} = 45 \cdots \cdots \left( 2 \right)
nCr+1=120⋯⋯(3){}^n{C_{r + 1}} = 120 \cdots \cdots \left( 3 \right)
Now we are going to use the property of combination which is given by nCrnCr−1=n−r+1r⋯⋯(4)\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} \cdots \cdots \left( 4 \right).
Now we are going to substitute values from equation (1)\left( 1 \right) and (2)\left( 2 \right) in equation (4)\left( 4 \right). Therefore, we get
4510=n−r+1r ⇒92=n−r+1r ⇒9r=2(n−r+1) ⇒9r=2n−2r+2 ⇒9r+2r−2n=2 ⇒11r−2n=2⋯⋯(5)  \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 9r + 2r - 2n = 2 \\\ \Rightarrow 11r - 2n = 2 \cdots \cdots \left( 5 \right) \\\
Let us replace rr by r+1r + 1 in equation (4)\left( 4 \right). Therefore, we get
nCr+1nCr+1−1=n−(r+1)+1r+1 ⇒nCr+1nCr=n−r−1+1r+1 ⇒nCr+1nCr=n−rr+1⋯⋯(6)  \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}} \cdots \cdots \left( 6 \right) \\\
Now we are going to substitute values from equation (2)\left( 2 \right) and (3)\left( 3 \right) in equation (6)\left( 6 \right). Therefore, we get
12045=n−rr+1 ⇒83=n−rr+1 ⇒8(r+1)=3(n−r) ⇒8r+8=3n−3r ⇒8r+3r−3n=−8 ⇒11r−3n=−8⋯⋯(7)  \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 8r + 3r - 3n = - 8 \\\ \Rightarrow 11r - 3n = - 8 \cdots \cdots \left( 7 \right) \\\
Now we have two equations in two variables nn and rr. To find the value of rr, we will eliminate nn from the equations (5)\left( 5 \right) and (7)\left( 7 \right). For this, we will multiply by number 33 on both sides of equation (5)\left( 5 \right) and also multiply by number 22 on both sides of equation (7)\left( 7 \right). Therefore, we get
33r−6n=6⋯⋯(8) 22r−6n=−16⋯⋯(9)  33r - 6n = 6 \cdots \cdots \left( 8 \right) \\\ 22r - 6n = - 16 \cdots \cdots \left( 9 \right) \\\
Now we will subtract equation (9)\left( 9 \right) from equation (8)\left( 8 \right). Therefore, we get
(33r−6n)−(22r−6n)=6−(−16) ⇒33r−6n−22r+6n=6+16 ⇒11r=22 ⇒r=2211 ⇒r=2  \left( {33r - 6n} \right) - \left( {22r - 6n} \right) = 6 - \left( { - 16} \right) \\\ \Rightarrow 33r - 6n - 22r + 6n = 6 + 16 \\\ \Rightarrow 11r = 22 \\\ \Rightarrow r = \dfrac{{22}}{{11}} \\\ \Rightarrow r = 2 \\\
Therefore, option B is correct.

Note: When solving these type of problems,make use of the appropriate property/formula of combination in accordance with the result which has to be found out/proved