Question

If $n$ is the natural number. Then, the range of the function $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$,$4 \leqslant n \leqslant 6$, is
A. \left\\{ {1,2,3,4} \right\\}
B. \left\\{ {1,2,3,4,5,6} \right\\}
C. \left\\{ {1,2,3} \right\\}
D. \left\\{ {1,2,3,4,5} \right\\}
E. $\phi$

Answer 1

In order to find the range of the function given $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$, we need to know about the conditions of the permutation given that will satisfy the equation and also check the conditions of general permutation. Then according to the overall conditions, get the domain, and substitute the value of domain in the equation and get the range of the equation accordingly.

Formula used:
$^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Complete step by step answer:
We are given with an equation $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$. We need to know about the domain of the function, in order to find the range. And for the domain, we need to satisfy all the conditions. From Permutation $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, we know some of the conditions as:
$n \geqslant 0$, $r \geqslant 0$ and $n \geqslant r$.

Comparing $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$ with $^n{P_r}$ and their conditions, we get: For $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$:
1st Condition:
$8 - n \geqslant 0$
Adding both the sides by $n$:
$8 - n + n \geqslant 0 + n \\\ \Rightarrow 8 \geqslant n \\\$
$\Rightarrow n \leqslant 8$ …..(1)
Similarly,
2nd Condition:
$n - 4 \geqslant 0$
Adding both the sides by $4$:
$n - 4 + 4 \geqslant 0 + 4 \\\ \Rightarrow n \geqslant 4 \\\$
$\Rightarrow 4 \leqslant n$ …..(2)

And, the 3rd condition:
$8 - n \geqslant n - 4$
Adding both the sides by $n$ and $4$, we get:
$8 - n + n + 4 \geqslant n - 4 + 4 + n$
$\Rightarrow 12 \geqslant 2n$
Dividing both the sides by $2$:
$\Rightarrow \dfrac{{12}}{2} \geqslant \dfrac{{2n}}{2}$
$\Rightarrow 6 \geqslant n$
$\Rightarrow n \leqslant 6$ ……(3)
From 1, 2 and 3, the overall condition becomes:
$4 \leqslant n \leqslant 6$
which is the required domain and was already given to us.

Now, we have the function $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$, we would substitute the values of the domain$4 \leqslant n \leqslant 6$ and will check the range.
For the value $4$:
$f\left( 4 \right){ = ^{8 - 4}}{P_{4 - 4}}{ = ^4}{P_0} = \dfrac{{4!}}{{\left( {4 - 0} \right)!}} = \dfrac{{4!}}{{4!}} = 1$
For the value $5$:
$f\left( 5 \right){ = ^{8 - 5}}{P_{5 - 4}}{ = ^3}{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = \dfrac{{3 \times 2!}}{{2!}} = 3$
For the value $6$:
$f\left( 6 \right){ = ^{8 - 6}}{P_{6 - 4}}{ = ^2}{P_2} = \dfrac{{2!}}{{\left( {2 - 2} \right)!}} = \dfrac{{2!}}{{0!}} = \dfrac{2}{1} = 2$
Therefore, for the domain $4 \leqslant n \leqslant 6$, the range is $1,3,2$ that can be written as \left\\{ {1,2,3} \right\\}. Hence, the range of the function $f\left( n \right){ = ^{8 - n}}{P_{n - 4}}$,$4 \leqslant n \leqslant 6$, is \left\\{ {1,2,3} \right\\}.

Hence, option C is correct.

Note: It’s important to check the conditions of the function to find the domain of the function, then opt for finding the range. Otherwise, it would lead to error. $^n{P_r}$ is the representation of permutation. Permutation is a way of arranging the objects in order. Order is a must factor in permutation. Factorial of $0! = 1$.