Answer

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Hint: Use the relation $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ to solve the terms $P_{r}^{5}$ and $P_{r-1}^{6}$. Now, form an equation in terms of ‘r’ and try to solve it further to get the value of ’r’.

Complete step-by-step answer:

Here we have to determine the value of ‘r’ if $P_{r}^{5}=2P_{r-1}^{6}$ ………………….. (i)

Now, we can use the identity of $P_{r}^{n}$ which is given as

$P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ …………….. (ii)

Where n! = 1.2.3.4………. n

Hence, Applying the identity (ii) with the equation (i), we get

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 6-\left( r-1 \right) \right)!}$

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 6-r+1 \right)!}$

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 7-r \right)!}$…………….. (iii)

Now, as we know expansion of n! can be given as n! = 1.2.3……………. n

So, we get

5! = 1.2.3.4.5 and 6! = 1.2.3.4.5.6

Now, let us write the expression of (7-r)! as following

(7-r)! = 1.2.3……… (7-r)

Let us write the above expression in reverse order, we get

(7-r)! = (7-r) (7-r-1) (7-r-2) ……………..3.2.1

Or

(7-r)! = (7-r) (6-r) (5-r) (4-r)……………3.2.1

Now, we can observe that the expansion (5-r) (4-r) (3-r)………..3.2.1 or 1.2.3…….. (3-r) (4-r) (5-r) can be replaced by (5-r)!.

Hence, we get

(7-r)! = (7-r) (6-r) (5-r)!............ (iv)

Now, putting the values of 5!, 6! and (7-r)! (from equation (iv)) in equation (iii), we get

$\dfrac{5\times 4\times 3\times 2\times 1}{\left( 5-r \right)!}=\dfrac{2\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 7-r \right)\left( 6-r \right)\left( 5-r \right)!}$

Now, cancelling out (5-r)! and $5\times 4\times 3\times 2\times 1$ from both the sides, we get

$\dfrac{1}{1}=\dfrac{2\times 6}{\left( 7-r \right)\left( 6-r \right)}$

On cross multiplication, we get

(7-r) (6-r) = 12

Now, we can observe that (7-r) and (6-r) have a difference of 1 and we can break 12 in 4 and 3. So, we can get equation as $(7-r)(6-r)=4\times 3$

Now, we know that $r\ge 0$ as ‘n’ and ‘r’ both should be positive in $p_{r}^{n}$ (‘n’ cannot be zero). And hence the value of ‘7-r’ is greater than ‘6-r’. So,

7-r=4 and 6-r=3

And from both the equations, we get that r=3.

Hence, for $P_{r}^{5}={{2}^{6}}{{P}_{r-1}}$ the value of r will be 3.

Note: One can apply the formula of $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$ in place of $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$, this is the general mistake by students in permutations and combinations.

One can solve equation (6-r) (7-r) = 12 by factoring as well. So, we get on simplifying,

$\begin{align}

& 42-6r-7r+{{r}^{2}}=12 \\

& {{r}^{2}}-13r+30=0 \\

\end{align}$

(r-3) (r-7) = 0

r=3 or r=7.

r=7 cannot be possible with the expression $P_{r}^{5}$ and $P_{r-1}^{6}$. Therefore, r = 3.

Complete step-by-step answer:

Here we have to determine the value of ‘r’ if $P_{r}^{5}=2P_{r-1}^{6}$ ………………….. (i)

Now, we can use the identity of $P_{r}^{n}$ which is given as

$P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ …………….. (ii)

Where n! = 1.2.3.4………. n

Hence, Applying the identity (ii) with the equation (i), we get

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 6-\left( r-1 \right) \right)!}$

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 6-r+1 \right)!}$

$\dfrac{5!}{\left( 5-r \right)!}=2\times \dfrac{6!}{\left( 7-r \right)!}$…………….. (iii)

Now, as we know expansion of n! can be given as n! = 1.2.3……………. n

So, we get

5! = 1.2.3.4.5 and 6! = 1.2.3.4.5.6

Now, let us write the expression of (7-r)! as following

(7-r)! = 1.2.3……… (7-r)

Let us write the above expression in reverse order, we get

(7-r)! = (7-r) (7-r-1) (7-r-2) ……………..3.2.1

Or

(7-r)! = (7-r) (6-r) (5-r) (4-r)……………3.2.1

Now, we can observe that the expansion (5-r) (4-r) (3-r)………..3.2.1 or 1.2.3…….. (3-r) (4-r) (5-r) can be replaced by (5-r)!.

Hence, we get

(7-r)! = (7-r) (6-r) (5-r)!............ (iv)

Now, putting the values of 5!, 6! and (7-r)! (from equation (iv)) in equation (iii), we get

$\dfrac{5\times 4\times 3\times 2\times 1}{\left( 5-r \right)!}=\dfrac{2\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 7-r \right)\left( 6-r \right)\left( 5-r \right)!}$

Now, cancelling out (5-r)! and $5\times 4\times 3\times 2\times 1$ from both the sides, we get

$\dfrac{1}{1}=\dfrac{2\times 6}{\left( 7-r \right)\left( 6-r \right)}$

On cross multiplication, we get

(7-r) (6-r) = 12

Now, we can observe that (7-r) and (6-r) have a difference of 1 and we can break 12 in 4 and 3. So, we can get equation as $(7-r)(6-r)=4\times 3$

Now, we know that $r\ge 0$ as ‘n’ and ‘r’ both should be positive in $p_{r}^{n}$ (‘n’ cannot be zero). And hence the value of ‘7-r’ is greater than ‘6-r’. So,

7-r=4 and 6-r=3

And from both the equations, we get that r=3.

Hence, for $P_{r}^{5}={{2}^{6}}{{P}_{r-1}}$ the value of r will be 3.

Note: One can apply the formula of $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$ in place of $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$, this is the general mistake by students in permutations and combinations.

One can solve equation (6-r) (7-r) = 12 by factoring as well. So, we get on simplifying,

$\begin{align}

& 42-6r-7r+{{r}^{2}}=12 \\

& {{r}^{2}}-13r+30=0 \\

\end{align}$

(r-3) (r-7) = 0

r=3 or r=7.

r=7 cannot be possible with the expression $P_{r}^{5}$ and $P_{r-1}^{6}$. Therefore, r = 3.

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