Question

If $^{10}{{P}_{r}}=5040$, find the value of r.

Answer 1

Hint: In questions like these, always remember that both the sides around the equal sign should be in factorial notation, to find the unknown value.

Complete step-by-step answer:

We know that $^{n}{{P}_{r}}=\dfrac{\left| \!{\underline {,

n ,}} \right. }{\left| \!{\underline {,

n-r ,}} \right. }$

It is the number of ways in which we can choose r different objects out of a set containing n different objects. Remember $r\le n$ always.

So, $^{10}{{P}_{r}}$ , it can be written as $\dfrac{\left| \!{\underline {,

10 ,}} \right. }{\left| \!{\underline {,

10-r ,}} \right. }$ .

In the question, it is given that $^{10}{{P}_{r}}=5040$

$\Rightarrow \dfrac{\left| \!{\underline {,

10 ,}} \right. }{\left| \!{\underline {,

10-r ,}} \right. }=5040$

To solve these types of questions, both sides should be in factorial notation.

So, 5040 has to be written in a factorial notation of a natural number.

For this, we have to use the hit and trial method. In this, we have to start from the factorial of the first natural number that is 1 till the factorial of the number, whose value is equal to 5040.

Now, when we solve in this way, we will get $\left| \!{\underline {,

7 ,}} \right. =5040$

$\Rightarrow \dfrac{\left| \!{\underline {,

10 ,}} \right. }{\left| \!{\underline {,

10-r ,}} \right. }=\left| \!{\underline {,

7 ,}} \right. $

We know that, $\left| \!{\underline {,

n ,}} \right. =n\times (n-1)\times (n-2)\times (n-3)\times ........\times 1$

Also, $\left| \!{\underline {,

n ,}} \right. $can be written as$n\times \left| \!{\underline {,

n-1 ,}} \right. $

$\Rightarrow \left| \!{\underline {,

n ,}} \right. =n\times \left| \!{\underline {,

n-1 ,}} \right. $

Similarly, $\left| \!{\underline {,

n ,}} \right. =n\times (n-1)\times \left| \!{\underline {,

n-2 ,}} \right. $

So, this pattern can be followed up to 1.

$\Rightarrow \left| \!{\underline {,

10 ,}} \right. =10\times 9\times 8\times \left| \!{\underline {,

7 ,}} \right. $

Hence, $\dfrac{\left| \!{\underline {,

10 ,}} \right. }{\left| \!{\underline {,

10-r ,}} \right. }=\left| \!{\underline {,

7 ,}} \right. $ , it can be written as below.

& \dfrac{10\times 9\times 8\times \left| \\!{\underline {\, 7 \,}} \right. }{\left| \\!{\underline {\, 10-r \,}} \right. }=\left| \\!{\underline {\, 7 \,}} \right. \\\ & \Rightarrow \dfrac{10\times 9\times 8\times \left| \\!{\underline {\, 7 \,}} \right. }{\left| \\!{\underline {\, 10-r \,}} \right. }=\left| \\!{\underline {\, 7 \,}} \right. \\\ & \Rightarrow 10\times 9\times 8=\left| \\!{\underline {\, 10-r \,}} \right. \\\ & \Rightarrow \left| \\!{\underline {\, 10-r \,}} \right. =720 \\\ \end{aligned}$$ 720 can be written as $\left| \\!{\underline {\, 6 \,}} \right. $ $\Rightarrow \left| \\!{\underline {\, 10-r \,}} \right. =\left| \\!{\underline {\, 6 \,}} \right. $ Hence, equating them, we get, $\begin{aligned} & 10-r=6 \\\ & \Rightarrow r=10-6 \\\ & \Rightarrow r=4 \\\ \end{aligned}$ Hence, the value of r is 4. Note: Here, while solving the factorial notation, we have to solve a large number of calculations. So, be careful while performing those calculations as there are chances to commit mistakes. Also, most of the time, you can just cross it by simply equating it with the other side. This will reduce the calculation time.