Question

Find the ordered pair (n ,r) if ${}^{n}{{P}_{r}}=30240$ and ${}^{n}{{C}_{r}}$=252
A. (12, 6)
B. (10, 5)
C. (9, 4)
D. (16, 7)

Answer 1

To solve this question, we should know the formulae for ${}^{n}{{P}_{r}}$ and ${}^{n}{{C}_{r}}$.
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We can infer from the two formulae that
$\begin{aligned} & {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ & {}^{n}{{C}_{r}}\times r!=\dfrac{n!}{\left( n-r \right)!} \\\ & {}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r! \\\ \end{aligned}$
Using this relation and prime factorisation, we can find the r. After finding the value of r, we can use the relation ${}^{n}{{P}_{r}}=30240$ to find the value of n.

Complete step-by-step answer:
The formulae used in the question are,
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\to \left( 1 \right)$
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\to \left( 2 \right)$
From the two equations, we can write that
$\begin{aligned} & {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ & {}^{n}{{C}_{r}}\times r!=\dfrac{n!}{\left( n-r \right)!} \\\ & {}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!\to \left( 3 \right) \\\ \end{aligned}$
Using the given values of ${}^{n}{{P}_{r}}\text{ and }{}^{n}{{C}_{r}}$in the equation-3, we get
${}^{n}{{P}_{r}}=30240,{}^{n}{{C}_{r}}=252$
$\begin{aligned} & 30240=252\times r! \\\ & r!=\dfrac{30240}{252} \\\ & r!=120 \\\ \end{aligned}$
To find the value of r, we have to do the factorisation of 120 and write the product as a product of consecutive numbers.
Instead of doing factorisation by steps of 2, we can do the factorisation in steps of 4, to reduce the iterations. That is
$\begin{aligned} & 4\left| \\!{\underline {\, 120 \,}} \right. \\\ & 2\left| \\!{\underline {\, 30 \,}} \right. \\\ & 3\left| \\!{\underline {\, 5 \,}} \right. \\\ & 5\left| \\!{\underline {\, 1 \,}} \right. \\\ \end{aligned}$
120 = 4$\times$2$\times$3$\times$5
Arranging them in an order, we get
120 = 5$\times$4$\times$3$\times$2$\times$1
So, r! = 120 = 5$\times$4$\times$3$\times$2$\times$1 = 5!
So, r = 5.
Given that ${}^{n}{{P}_{r}}=30240$ using the value of r = 5 in the equation, we get
$\begin{aligned} & {}^{n}{{P}_{5}}=30240 \\\ & \dfrac{n!}{\left( n-5 \right)!}=30240 \\\ \end{aligned}$
Cancelling the common terms in the numerator and denominator, we get
$\begin{aligned} & \dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times \left( n-4 \right)\times \left( n-5 \right)!}{\left( n-5 \right)!}=30240 \\\ & n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times \left( n-4 \right)=30240\to \left( 4 \right) \\\ \end{aligned}$
We have to do the factorisation of 30240 and arrange them in a product of 5 consecutive numbers to get the value of n. Doing the prime factorisation in steps of 8 and 9, we get
$\begin{aligned} & 8\left| \\!{\underline {\, 30240 \,}} \right. \\\ & 4\left| \\!{\underline {\, 3780 \,}} \right. \\\ & 9\left| \\!{\underline {\, 945 \,}} \right. \\\ & 5\left| \\!{\underline {\, 105 \,}} \right. \\\ & 3\left| \\!{\underline {\, 21 \,}} \right. \\\ & 7\left| \\!{\underline {\, 7 \,}} \right. \\\ & \left| \\!{\underline {\, 1 \,}} \right. \\\ \end{aligned}$
$\begin{aligned} & 30240=8\times 4\times 9\times 5\times 3\times 7 \\\ & 30240=9\times 8\times 7\times 4\times 5\times 3 \\\ & 30240=9\times 8\times 7\times \left( 2\times 5 \right)\times \left( 2\times 3 \right) \\\ & 30240=10\times 9\times 8\times 7\times 6 \\\ \end{aligned}$
Comparing the result with equation-4, we get
n = 10.
$\therefore$The ordered pair (n , r) = (10 , 5).

So, the correct answer is “Option B”.

Note: The calculation required to get the value of r is relatively easy compared to the calculation required to calculate the value of n. If we observe the options, there is only one option with the value of r = 5, which is option-B. So, after the calculation of the value of r, we can conclude the correct answer without further calculations. In this type of questions, the questioner generally will give some help to the students by giving only one option which suits the value of r. The main concept in the question is ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$, and this will be tested. If there are multiple options with the calculated value of r, we have to go by the process mentioned in the solution.