Question

If $P\left( n,r \right)=1680$ and $C\left( n,r \right)=70$ , then $69n+r!$ is equal to
A. $128$
B. $576$
C. $256$
D. $625$
E. $1152$

Answer 1

Hint: At first, we write the formula of permutation as $\dfrac{n!}{\left( n-r \right)!}$ and equate it to $1680$ . Then, we write the formula of the combination as $\dfrac{n!}{r!\left( n-r \right)!}$ and equate it to $70$ . Dividing the two equations will give us the value of r and then putting this value in one of the equations, gives a polynomial in n. Drawing the graph of the polynomial gives its roots.

Complete step-by-step solution:
In our academics, we have come across a chapter in maths, which was called arrangements. In that chapter, we had studied the number of ways to group and arrange some items or objects among themselves. The terms associated with them are called permutations and combinations. Permutation refers to selection and arranging the selection among themselves, whereas combination refers to selection alone.
The permutation of r objects out of n objects is denoted in two ways, which are ${}^{n}{{P}_{r}}$ and $P\left( n,r \right)$ . The corresponding formula for the number of permutations possible is given by,
$P\left( n,r \right)=\dfrac{n!}{\left( n-r \right)!}$
Similarly, the permutation of r objects out of n objects is denoted in two ways, which are ${}^{n}{{C}_{r}}$ and $C\left( n,r \right)$ . The corresponding formula for the number of permutations possible is given by,
$C\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}$
In this problem, we are given that,
$\begin{align}
  & P\left( n,r \right)=1680 \\
 & \Rightarrow \dfrac{n!}{\left( n-r \right)!}=1680....\left( i \right) \\
\end{align}$
Also,
$\begin{align}
  & C\left( n,r \right)=70 \\
 & \Rightarrow \dfrac{n!}{r!\left( n-r \right)!}=70....\left( ii \right) \\
\end{align}$
Dividing equation (i) by (ii), we get,
$\begin{align}
  & \Rightarrow \dfrac{n!}{\left( n-r \right)!}\times \dfrac{r!\left( n-r \right)!}{n!}=\dfrac{1680}{70} \\
 & \Rightarrow r!=24 \\
\end{align}$
Now, we know that the factorial of a number is unique and the factorial of $4$ is $24$ . So,
$\Rightarrow r=4$
Putting this value in equation (i),
$\begin{align}
  & \Rightarrow \dfrac{n!}{\left( n-4 \right)!}=1680 \\
 & \Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=1680 \\
\end{align}$
We will now find the value of n by trial-and-error method and checking for which value, the expression gives $1680$ . By any means, n must be greater than r for the permutation to occur. So, taking $n=5$,
$\Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=5\times 4\times 3\times 2=120$
Taking $n=6$ ,
$\Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=6\times 5\times 4\times 3=360$
Taking $n=7$ ,
$\Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=7\times 6\times 5\times 4=840$
Taking $n=8$ ,
$\Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=8\times 7\times 6\times 5=1680$
Since we got $1680$, the value of $n$ is $8$.
Now,
$69n+r!=\left( 69\times 8 \right)+24=576$
Thus, we can conclude that the value of $69n+r!$ will be $576$ which is option B.

Note: We should know the factorial of some numbers at least up to $10$ . Solving problems will be easy then. Also, we can solve it in another way. We get an equation in n. The roots of the polynomial will be known from its graph, drawn in some online graphing calculator like Geogebra. It will take less time.