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Question: Find the value of \(^r{C_5}\) if \(^{18}{C_r}{ = ^{18}}{C_{r + 2}}\)....

Find the value of rC5^r{C_5} if 18Cr=18Cr+2^{18}{C_r}{ = ^{18}}{C_{r + 2}}.

Explanation

Solution

Hint: The formula for nCr^n{C_r} is n!r!(n−r)!\dfrac{{n!}}{{r!\left( {n - r} \right)!}}. Use this formula to find out the value of rr. And then put the value of rr in rC5^r{C_5}.

Complete step-by-step answer:
According to the question, it is given that 18Cr=18Cr+2^{18}{C_r}{ = ^{18}}{C_{r + 2}}.
We know that, nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}. Using this formula we’ll get:

⇒18Cr=18Cr+2 ⇒18!r!(18−r)!=18!(r+2)!(18−r−2)! ⇒1r!(18−r)!=1(r+2)!(16−r)! ⇒r!(18−r)!=(r+2)!(16−r)!  { \Rightarrow ^{18}}{C_r}{ = ^{18}}{C_{r + 2}} \\\ \Rightarrow \dfrac{{18!}}{{r!\left( {18 - r} \right)!}} = \dfrac{{18!}}{{\left( {r + 2} \right)!\left( {18 - r - 2} \right)!}} \\\ \Rightarrow \dfrac{1}{{r!\left( {18 - r} \right)!}} = \dfrac{1}{{\left( {r + 2} \right)!\left( {16 - r} \right)!}} \\\ \Rightarrow r!\left( {18 - r} \right)! = \left( {r + 2} \right)!\left( {16 - r} \right)! \\\

We know that, n!=n(n−1)(n−2).....3×2×1n! = n\left( {n - 1} \right)\left( {n - 2} \right).....3 \times 2 \times 1. Using this, we’ll get:

⇒r!(18−r)(18−r−1)(18−r−2)!=(r+2)(r+1)r!(16−r)! ⇒(18−r)(17−r)(16−r)!=(r+2)(r+1)(16−r)! ⇒(18−r)(17−r)=(r+2)(r+1) ⇒306−18r−17r+r2=r2+r+2r+2 ⇒306−35r=3r+2 ⇒38r=304 ⇒r=8  \Rightarrow r!\left( {18 - r} \right)\left( {18 - r - 1} \right)\left( {18 - r - 2} \right)! = \left( {r + 2} \right)\left( {r + 1} \right)r!\left( {16 - r} \right)! \\\ \Rightarrow \left( {18 - r} \right)\left( {17 - r} \right)\left( {16 - r} \right)! = \left( {r + 2} \right)\left( {r + 1} \right)\left( {16 - r} \right)! \\\ \Rightarrow \left( {18 - r} \right)\left( {17 - r} \right) = \left( {r + 2} \right)\left( {r + 1} \right) \\\ \Rightarrow 306 - 18r - 17r + {r^2} = {r^2} + r + 2r + 2 \\\ \Rightarrow 306 - 35r = 3r + 2 \\\ \Rightarrow 38r = 304 \\\ \Rightarrow r = 8 \\\

So, the value of rr is 8.

We have to find out the value of rC5^r{C_5}. Putting r=8r = 8, we’ll get:
⇒rC5=8C5{ \Rightarrow ^r}{C_5}{ = ^8}{C_5}
Using formula nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we’ll get:
⇒8C5=8!5!×3! ⇒8C5=8×7×6×5!5!×6 ⇒8C5=56  { \Rightarrow ^8}{C_5} = \dfrac{{8!}}{{5! \times 3!}} \\\ { \Rightarrow ^8}{C_5} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5! \times 6}} \\\ { \Rightarrow ^8}{C_5} = 56 \\\
Thus, the value of rC5^r{C_5} is 56.

Note: This question can be solved by another method as:
We know that if nCa=nCb^n{C_a}{ = ^n}{C_b} then either a=ba = b or a+b=na + b = n must be true.
So for 18Cr=18Cr+2^{18}{C_r}{ = ^{18}}{C_{r + 2}}, we have:
⇒r=r+2 or r+r+2=18\Rightarrow r = r + 2{\text{ or }}r + r + 2 = 18
First condition is not true. So we have:
⇒r+r+2=18 ⇒2r+2=18 ⇒2r=16 ⇒r=8  \Rightarrow r + r + 2 = 18 \\\ \Rightarrow 2r + 2 = 18 \\\ \Rightarrow 2r = 16 \\\ \Rightarrow r = 8 \\\
We have calculated the value of rr. While putting it in rC5^r{C_5} we will get the same result.