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# Let r and n be positive integers such that $1 \leqslant r \leqslant n$. Then prove the following:$\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$

Last updated date: 26th Mar 2023
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Answer
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Hint: We are given, in this question, a ratio of two combinations. We have to solve this ratio to reach the RHS part of the equation $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$. We do this by writing the values of the given combinations in factorial form. After doing so, we will try to cancel out most of the factorial values. In this way we will prove this equation.
Formula used: The method of collecting $r$ objects from a set of $n$ objects, where order of selection does not matter is given as ${}^n{C_r}$ and
${}^n{C_r} = \dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, r \,}} \right. }}$
For any factorial$\left| \!{\underline {\, a \,}} \right.$, we can also write it as,
$\left| \!{\underline {\, a \,}} \right. = a\left| \!{\underline {\, {a - 1} \,}} \right.$.

Complete step-by-step solution:
Here, we are given a ratio of two combinations$\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}$. We know that since $1 \leqslant r \leqslant n$, we can write,
${}^n{C_r} = \dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, r \,}} \right. }}$ Where $\left| \!{\underline {\, {} \,}} \right.$ means factorial and
${}^n{C_{r - 1}} = \dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r + 1} \,}} \right. \left| \!{\underline {\, {r - 1} \,}} \right. }}$.
On dividing both of them as per the LHS, we get,
$\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{\dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, r \,}} \right. }}}}{{\dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r + 1} \,}} \right. \left| \!{\underline {\, {r - 1} \,}} \right. }}}} \\ \Rightarrow \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{\left| \!{\underline {\, n \,}} \right. \left| \!{\underline {\, {n - r + 1} \,}} \right. \left| \!{\underline {\, {r - 1} \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, r \,}} \right. \left| \!{\underline {\, n \,}} \right. }} \\ \Rightarrow \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{\left| \!{\underline {\, {n - r + 1} \,}} \right. \left| \!{\underline {\, {r - 1} \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, r \,}} \right. }} \\$
Now we know that for any factorial $\left| \!{\underline {\, a \,}} \right.$, we can also write it as, $\left| \!{\underline {\, a \,}} \right. = a\left| \!{\underline {\, {a - 1} \,}} \right.$,
Using this we move ahead as,
$\Rightarrow \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{(n - r + 1)\left| \!{\underline {\, {n - r} \,}} \right. \left| \!{\underline {\, {r - 1} \,}} \right. }}{{r\left| \!{\underline {\, {r - 1} \,}} \right. \left| \!{\underline {\, {n - r} \,}} \right. }}$
On simplifying further,
$\Rightarrow \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{(n - r + 1)}}{r}$
This is the same as the expression we wanted to prove. Hence we have proved that
$\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{(n - r + 1)}}{r}$

Note: The concept of combination is very important and highly used in Binomial theorem and Probability theory. With the help of it we can simplify the complex calculations in both Binomial theorem and Probability theory. The above equation we have proved is very important and very useful as with the help of it we can come back from the combination terms to the normal terms.