Question

If \[{}^8P_r = {}^7\operatorname{P}_ 4 + 4.{}^7\operatorname{P}_3\] ,then \[{\text{r}} = ?\]
A) 6
B) 5
C) 7
D) 4

Answer 1

Hint: We are given with an equation, we will directly apply the formula of permutation \[{}^{\text{n}}P_r \] on three of the terms and solve the equation further to get the value of $r$.

Formula Used:
For permutation \[{}^{\text{n}}P_r \] is given by:
\[{}^{\text{n}}P_r = \dfrac{{{\text{n!}}}}{{\left( {{\text{n}} - {\text{r}}} \right){\text{!}}}}\]

Complete step by step solution:
The given equation is :
\[{}^8P_r = {}^7\operatorname{P}_4 + 4.{}^7\operatorname{P}_3..........................\left( 1 \right)\].
Now in order to get the answer for $r$ we will apply the following formula for three of the terms in the equation:
\[{}^{\text{n}}P_r = \dfrac{{{\text{n!}}}}{{\left( {{\text{n}} - {\text{r}}} \right){\text{!}}}}\]
Applying this formula for \[{}^8P_r \] we get:-
\[{}^8P_r = \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}}\]
Applying this formula for \[{}^7\operatorname{P}_4\] we get:-
\[
  {}^7\operatorname{P}_4 = \dfrac{{{\text{7!}}}}{{\left( {7 - 4} \right){\text{!}}}} \\
  {}^7\operatorname{P}_4 = \dfrac{{7!}}{{3!}} \\
 \]
Applying the formula for \[{}^7\operatorname{P}_3\] we get:-
\[
  {}^7\operatorname{P}_3 = \dfrac{{{\text{7!}}}}{{\left( {7 - 3} \right){\text{!}}}} \\
  {}^7\operatorname{P}_3 = \dfrac{{7!}}{{4!}} \\
 \]
Now putting the respective values of \[{}^8P_r \] , \[{}^7\operatorname{P}_4\] and \[{}^7\operatorname{P}_3\] in equation 1 we get:-
\[{}^8P_r = {}^7\operatorname{P}_4 + 4.{}^7\operatorname{P}_3\]
\[
  \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}} = \dfrac{{7!}}{{3!}} + 4\left( {\dfrac{{7!}}{{4!}}} \right) \\
  \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}} = \dfrac{{7!}}{{3!}} + 4\left( {\dfrac{{7!}}{{4 \times 3!}}} \right) \\
  \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}} = \dfrac{{7!}}{{3!}} + \left( {\dfrac{{7!}}{{3!}}} \right) \\
  \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}} = 2\left( {\dfrac{{7!}}{{3!}}} \right) \\
  \dfrac{{{\text{8}} \times {\text{7!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}} = 2\left( {\dfrac{{7!}}{{3 \times 2 \times 1}}} \right) \\
  \dfrac{8}{{\left( {8 - {\text{r}}} \right)!}} = \dfrac{1}{3} \\
 \]
Now cross multiplying both the sides and solving this equation further we get:-
\[
  8 \times 3 = \left( {8 - {\text{r}}} \right)! \\
  \left( {8 - {\text{r}}} \right)! = 24 \\
 \]
Now since we know,
\[
  4! = 4 \times 3 \times 2 \times 1 \\
  4! = 24 \\
 \]
Therefore,
\[
  8 - r = 4 \\
  8 - 4 = r \\
  r = 4 \\
 \]

The value of $r$ is 4. Therefore, option D is correct.

Note:
In this question, we do not have to solve for the proper values of right-hand side terms.
Also, the formula for \[{}^{\text{n}}Pr \] is:
\[{}^{\text{n}}Pr = \dfrac{{{\text{n!}}}}{{\left( {{\text{n}} - {\text{r}}} \right){\text{!}}}}\]
A permutation is used when we have arranged the things in a row i.e, it is used for arrangement.