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

If 8Pr=7P⁡4+4.7P⁡3{}^8P_r = {}^7\operatorname{P}_ 4 + 4.{}^7\operatorname{P}_3 ,then r=?{\text{r}} = ?
A) 6
B) 5
C) 7
D) 4

Explanation

Solution

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

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

Complete step by step solution:
The given equation is :
8Pr=7P⁡4+4.7P⁡3..........................(1){}^8P_r = {}^7\operatorname{P}_4 + 4.{}^7\operatorname{P}_3..........................\left( 1 \right).
Now in order to get the answer for rr we will apply the following formula for three of the terms in the equation:
nPr=n!(n−r)!{}^{\text{n}}P_r = \dfrac{{{\text{n!}}}}{{\left( {{\text{n}} - {\text{r}}} \right){\text{!}}}}
Applying this formula for 8Pr{}^8P_r we get:-
8Pr=8!(8−r)!{}^8P_r = \dfrac{{{\text{8!}}}}{{\left( {8 - {\text{r}}} \right){\text{!}}}}
Applying this formula for 7P⁡4{}^7\operatorname{P}_4 we get:-

7P⁡4=7!(7−4)! 7P⁡4=7!3!  {}^7\operatorname{P}_4 = \dfrac{{{\text{7!}}}}{{\left( {7 - 4} \right){\text{!}}}} \\\ {}^7\operatorname{P}_4 = \dfrac{{7!}}{{3!}} \\\

Applying the formula for 7P⁡3{}^7\operatorname{P}_3 we get:-

7P⁡3=7!(7−3)! 7P⁡3=7!4!  {}^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 8Pr{}^8P_r , 7P⁡4{}^7\operatorname{P}_4 and 7P⁡3{}^7\operatorname{P}_3 in equation 1 we get:-
8Pr=7P⁡4+4.7P⁡3{}^8P_r = {}^7\operatorname{P}_4 + 4.{}^7\operatorname{P}_3

8!(8−r)!=7!3!+4(7!4!) 8!(8−r)!=7!3!+4(7!4×3!) 8!(8−r)!=7!3!+(7!3!) 8!(8−r)!=2(7!3!) 8×7!(8−r)!=2(7!3×2×1) 8(8−r)!=13  \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:-

8×3=(8−r)! (8−r)!=24  8 \times 3 = \left( {8 - {\text{r}}} \right)! \\\ \left( {8 - {\text{r}}} \right)! = 24 \\\

Now since we know,

4!=4×3×2×1 4!=24  4! = 4 \times 3 \times 2 \times 1 \\\ 4! = 24 \\\

Therefore,

8−r=4 8−4=r r=4  8 - r = 4 \\\ 8 - 4 = r \\\ r = 4 \\\

The value of rr 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 nPr{}^{\text{n}}Pr is:
nPr=n!(n−r)!{}^{\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.