# If ${}^n{P_r} = {}^n{P_{r + 1}}$ and ${}^n{C_r} = {}^n{C_{r - 1}}$ , find n and r.

Last updated date: 26th Mar 2023

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Answer

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Hint: By the use of Permutation and Combination formulae and properties we will find the values of n and r.

Given,

${}^n{P_r} = {}^n{P_{r + 1}} \to (1)$

Since, we know that the number of permutations of n objects taken r at a time is ${}^n{P_r}$ and its value is $\dfrac{{n!}}{{(n - r)!}}$ . Similarly, the value of ${}^n{P_{r + 1}}$ is equal to $\dfrac{{n!}}{{(n - r - 1)!}}$. So, applying the formulae in equation (1), we get

$

\Rightarrow \dfrac{{n!}}{{(n - r)!}} = \dfrac{{n!}}{{(n - r - 1)!}} \\

\Rightarrow (n - r - 1)! = (n - r)! \to (2) \\

$

As, we know that $n! = n*(n - 1)!$ . Therefore, $(n - r)! = (n - r)*(n - r - 1)!$ . Rewriting equation (2), we get

$ \Rightarrow (n - r - 1)! = (n - r)*(n - r - 1)!$

The term $(n - r - 1)!$ gets cancelled on both sides, then the equation will be

$

\Rightarrow 1 = (n - r) \\

\Rightarrow n = r + 1 \to (3) \\

$

Now, let us consider the given condition ${}^n{C_r} = {}^n{C_{r - 1}} \to (4)$. As, we know the property of combination i.e.., if ${}^n{C_x} = {}^n{C_y}$ then $x = y$ or $x + y = n$. Therefore, using the property of combinations, equation (4) can be written as

$

\Rightarrow r + r - 1 = n \\

\Rightarrow 2r - 1 = n \\

$

Let us substitute the value of n from equation (3) in the above equation, we get

$

\Rightarrow 2r - 1 = r + 1 \\

\Rightarrow 2r - r = 1 + 1 \\

\Rightarrow r = 2 \\

$

Hence, the obtained value of r is 2.let us substitute the value of r in equation (3), we get

$

\Rightarrow n = 2 + 1 \\

\Rightarrow n = 3 \\

$

Therefore, the obtained value of n is 3 and r is 2.

Note: Here, we have considered $r + r - 1 = n$ condition, because if we consider the condition $r = r - 1$ the value of r can’t be computed.

Given,

${}^n{P_r} = {}^n{P_{r + 1}} \to (1)$

Since, we know that the number of permutations of n objects taken r at a time is ${}^n{P_r}$ and its value is $\dfrac{{n!}}{{(n - r)!}}$ . Similarly, the value of ${}^n{P_{r + 1}}$ is equal to $\dfrac{{n!}}{{(n - r - 1)!}}$. So, applying the formulae in equation (1), we get

$

\Rightarrow \dfrac{{n!}}{{(n - r)!}} = \dfrac{{n!}}{{(n - r - 1)!}} \\

\Rightarrow (n - r - 1)! = (n - r)! \to (2) \\

$

As, we know that $n! = n*(n - 1)!$ . Therefore, $(n - r)! = (n - r)*(n - r - 1)!$ . Rewriting equation (2), we get

$ \Rightarrow (n - r - 1)! = (n - r)*(n - r - 1)!$

The term $(n - r - 1)!$ gets cancelled on both sides, then the equation will be

$

\Rightarrow 1 = (n - r) \\

\Rightarrow n = r + 1 \to (3) \\

$

Now, let us consider the given condition ${}^n{C_r} = {}^n{C_{r - 1}} \to (4)$. As, we know the property of combination i.e.., if ${}^n{C_x} = {}^n{C_y}$ then $x = y$ or $x + y = n$. Therefore, using the property of combinations, equation (4) can be written as

$

\Rightarrow r + r - 1 = n \\

\Rightarrow 2r - 1 = n \\

$

Let us substitute the value of n from equation (3) in the above equation, we get

$

\Rightarrow 2r - 1 = r + 1 \\

\Rightarrow 2r - r = 1 + 1 \\

\Rightarrow r = 2 \\

$

Hence, the obtained value of r is 2.let us substitute the value of r in equation (3), we get

$

\Rightarrow n = 2 + 1 \\

\Rightarrow n = 3 \\

$

Therefore, the obtained value of n is 3 and r is 2.

Note: Here, we have considered $r + r - 1 = n$ condition, because if we consider the condition $r = r - 1$ the value of r can’t be computed.

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