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Question: Find the value of n If \[10\ ^{n} C_{2}=3\ ^{n+1} C_{3}\]....

Find the value of n If 10 nC2=3 n+1C310\ ^{n} C_{2}=3\ ^{n+1} C_{3}.

Explanation

Solution

Hint: In this question it is given that if 10 nC2=3 n+1C310\ ^{n} C_{2}=3\ ^{n+1} C_{3} then we have to find the value of n. So to find the solution we have to use the combination formula,
i.e, nCr=n!r!⋅(n−r)!{}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}......(1)
Where, n!=n⋅(n−1)⋅(n−2)⋯3⋅2⋅1n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 3\cdot 2\cdot 1 and also we can write n!=n⋅(n−1)!n!=n\cdot \left( n-1\right) !
Complete step-by-step solution:
Given equation,
10 nC2=3 n+1C310\ ^{n} C_{2}=3\ ^{n+1} C_{3}
⇒10 n!2!⋅(n−2)!=3 (n+1)!3!⋅(n+1−3)!\Rightarrow 10\ \dfrac{n!}{2!\cdot \left( n-2\right) !} =3\ \dfrac{\left( n+1\right) !}{3!\cdot \left( n+1-3\right) !}
⇒10 n!2⋅1⋅(n−2)!=3 (n+1)⋅n!3⋅2⋅1⋅(n−2)!\Rightarrow 10\ \dfrac{n!}{2\cdot 1\cdot \left( n-2\right) !} =3\ \dfrac{\left( n+1\right) \cdot n!}{3\cdot 2\cdot 1\cdot \left( n-2\right) !}
⇒10 n!2⋅1⋅(n−2)!= (n+1)⋅n!2⋅1⋅(n−2)!\Rightarrow 10\ \dfrac{n!}{2\cdot 1\cdot \left( n-2\right) !} =\ \dfrac{\left( n+1\right) \cdot n!}{2\cdot 1\cdot \left( n-2\right) !}
⇒10 n!(n−2)!= (n+1)⋅n!(n−2)!\Rightarrow 10\ \dfrac{n!}{\left( n-2\right) !} =\ \dfrac{\left( n+1\right) \cdot n!}{\left( n-2\right) !}
⇒10⋅n!= (n+1)⋅n!\Rightarrow 10\cdot n!=\ \left( n+1\right) \cdot n! [ cancelling (n-2)! on the both side of denominator]
⇒10=(n+1)\Rightarrow 10=\left( n+1\right) [canceling n! on the both side]
⇒(n+1)=10\Rightarrow \left( n+1\right) =10
⇒n=10−1\Rightarrow n=10-1
⇒n=9\Rightarrow n=9
Note: While solving this type of problem you need to know that nCr{}^{n}C_{r} defines choosing r number of different items from n number of different items also to solve the combination related equation you have to expand the equation upto a certain steps, like we expand during the solution.