Question

If ${}^{5}{{P}_{r}}=120$, then the value of r is
A. 2
B. 3+
C. 5
D. 1

Answer 1

The given question is a one based on permutation. To find the value of r, we will solve the given condition with respect to r. We know that the formula of permutation is ${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$. Substituting the values given in the question to this formula we will have the expression in terms of r. Then solving the expression to find a value of r that satisfies the given condition.

Complete step by step solution:
Permutation is a mathematical concept that describes the number of possibilities or the number of ways objects can be arranged. It involves the use of order to predict the possibilities unlike the combination concept.
Mathematically, permutation can be written as:
${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$
where n = total number of objects
r = number of selected objects
and ${}^{n}{{P}_{r}}$ is the permutation of the selected objects with respect to total number of objects.
According to the question, we have been given a permutation and we have to find the value of r based on that given permutation,
We have,
${}^{5}{{P}_{r}}=120$
Applying the permutation formula, we get,
$\Rightarrow \dfrac{5!}{(5-r)!}=120$--------(1)
As we know that, $5!=5\times 4\times 3\times 2\times 1=120$
We have $120$ on either side of the equality so it is cancelled, that is,
$\Rightarrow \dfrac{5\times 4\times 3\times 2\times 1}{(5-r)!}=120$
$\Rightarrow \dfrac{120}{(5-r)!}=120$
$\Rightarrow \dfrac{1}{(5-r)!}=1$
We now have,
$\Rightarrow (5-r)!=1$---------(2)
We have to put a value of r so that the above equation is satisfied, we have,
$r=5$
We took $r=5$ because if we substitute the value of $r=5$ in the equation (2) , we get,
$(5-r)!$
Putting $r=5$, we get,
$(5-5)!=0!=1$ which satisfies condition in equation (2)
Therefore, the value of $r=5$.

So, the correct answer is “Option C”.

Note: The formula of permutation should not be confused with the formula for combination as the formula for both is almost the same except for a small variation.
${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$ is the formula for permutation and that for combination is ${}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$.
On dividing we get,
$\dfrac{{}^{n}{{P}_{r}}}{{}^{n}{{C}_{r}}}=\dfrac{\dfrac{n!}{(n-r)!}}{\dfrac{n!}{(n-r)!r!}}=r!$
That is, ${}^{n}{{P}_{r}}=r!{}^{n}{{C}_{r}}$ or ${}^{n}{{C}_{r}}=\dfrac{{}^{n}{{P}_{r}}}{r!}$
The difference between the two formula is of $r!$