Question

If $^{n}{{P}_{r}}$ = 3024 then (n, r) is equal to.
$\begin{align}
  & a)\left( 8,4 \right) \\
 & b)\left( 8,3 \right) \\
 & c)\left( 9,3 \right) \\
 & d)\left( 9,4 \right) \\
\end{align}$

Answer 1

Hint: Now we know that $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ Now since we have $^{n}{{P}_{r}}$ = 3024 we will check the values of n and r for which the equation is true.

Complete step by step answer:
Now let us understand what is $^{n}{{P}_{r}}$ .
Here P means Permutation. Permutation is nothing but arrangement of objects.
Now consider we haven blue balls and we want to arrange them in r places.
Then the total number of possible arrangements of n balls into r places is given by the number $^{n}{{P}_{r}}$
Now let us see this with an example.
If we have 4 balls and we have 2 boxes. Then the number of ways to arrange them is given by $^{4}{{P}_{2}}$
Now for any n and r the number $^{n}{{P}_{r}}$ is given by $\dfrac{n!}{\left( n-r \right)!}$ where $a!=a\times \left( a-1 \right)\times \left( a-2 \right)\times ....\times 3\times 2\times 1$
Now consider our same example where we have 4 balls and we want to arrange them in 2 boxes.
Then the total number of ways to do this is given by
$^{4}{{P}_{2}}=\dfrac{4!}{\left( 4-2 \right)!}$
Now using definition of factorial we get
$^{4}{{P}_{2}}=\dfrac{4\times 3\times 2\times 1}{2\times 1}=4\times 3=12$
Hence we get there are 12 possible arrangements.
Now let us check this by counting also
Since we have 4 balls let us say ball 1, ball 2, ball 3. Ball 4.
Then if we place ball 1 in box 1 then in other box we have 3 options
Similarly if we place ball 2 in box 1 then in other box we have 3 options
Same goes for ball 3 and ball 4.
Hence total number of arrangements we get = 3 + 3 + 3 + 3 = 12.
Now we have $^{n}{{P}_{r}}$ = 3024
Now using formula we get
$\dfrac{n!}{\left( n-r \right)!}=3024$
Now for n = 8 r = 4 we get
$^{8}{{P}_{4}}=\dfrac{8!}{\left( 8-4 \right)!}=\dfrac{8!}{4!}=\dfrac{8\times 7\times 6\times 5\times 4!}{4!}=8\times 7\times 6\times 5=1680$
Now for n = 8 r = 3
$^{8}{{P}_{4}}=\dfrac{8!}{\left( 8-3 \right)!}=\dfrac{8!}{3!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3!}{3!}=8\times 7\times 6\times 5\times 4=6720$
Now for n = 9 r = 3
$^{9}{{P}_{3}}=\dfrac{9!}{\left( 9-3 \right)!}=\dfrac{9!}{6!}=\dfrac{9\times 8\times 7\times 6!}{6!}=9\times 8\times 7=504$
Now for n = 9 r = 4
$^{9}{{P}_{4}}=\dfrac{9!}{\left( 9-4 \right)!}=\dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times 5!}{5!}=9\times 8\times 7\times 6=3024$
Hence we have n = 9 and r = 4.

So, the correct answer is “Option D”.

Note: Note that permutation and combination are different. Permutations is number of possible arrangements while combination is number of possible selections and $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$