Question

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

Answer 1

Hint: In this question it is given that if $$10\ ^{n} C_{2}=3\ ^{n+1} C_{3}$$ then we have to find the value of n. So to find the solution we have to use the combination formula,
i.e, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$......(1)
Where, $$n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 3\cdot 2\cdot 1$$ and also we can write $$n!=n\cdot \left( n-1\right) !$$
Complete step-by-step solution:
Given equation,
$$10\ ^{n} C_{2}=3\ ^{n+1} C_{3}$$
$$\Rightarrow 10\ \dfrac{n!}{2!\cdot \left( n-2\right) !} =3\ \dfrac{\left( n+1\right) !}{3!\cdot \left( n+1-3\right) !}$$
$$\Rightarrow 10\ \dfrac{n!}{2\cdot 1\cdot \left( n-2\right) !} =3\ \dfrac{\left( n+1\right) \cdot n!}{3\cdot 2\cdot 1\cdot \left( n-2\right) !}$$
$$\Rightarrow 10\ \dfrac{n!}{2\cdot 1\cdot \left( n-2\right) !} =\ \dfrac{\left( n+1\right) \cdot n!}{2\cdot 1\cdot \left( n-2\right) !}$$
$$\Rightarrow 10\ \dfrac{n!}{\left( n-2\right) !} =\ \dfrac{\left( n+1\right) \cdot n!}{\left( n-2\right) !}$$
$$\Rightarrow 10\cdot n!=\ \left( n+1\right) \cdot n!$$ [ cancelling (n-2)! on the both side of denominator]
$$\Rightarrow 10=\left( n+1\right) $$ [canceling n! on the both side]
$$\Rightarrow \left( n+1\right) =10$$
$$\Rightarrow n=10-1$$
$$\Rightarrow n=9$$
Note: While solving this type of problem you need to know that ${}^{n}C_{r}$ 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.