If 8 cdot {}^{7}P_{r} = 7 cdot {}^{8}P_{r-1}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation: $8 \cdot {}^{7}P_{r} = 7 \cdot {}^{8}P_{r-1}$ Using the formula ${}^{n}P_{r} = \frac{n!}{(n-r)!}$: $8 \cdot \frac{7!}{(7-r)!} = 7

Vedclass · Accepted Answer

No solution (No such $r$ exists)

Vedclass · Answer

(D) Given equation: $8 \cdot {}^{7}P_{r} = 7 \cdot {}^{8}P_{r-1}$ Using the formula ${}^{n}P_{r} = \frac{n!}{(n-r)!}$: $8 \cdot \frac{7!}{(7-r)!} = 7 \cdot \frac{8!}{(8-(r-1))!} $ $8 \cdot \frac{7!}{(7-r)!} = 7 \cdot \frac{8!}{(9-r)!} $ Since $8! = 8 \cdot 7!$,we have: $\frac{8 \cdot 7!}{(7-r)!} = \frac{7 \cdot 8 \cdot 7!}{(9-r)!} $ $\frac{1}{(7-r)!} = \frac{7}{(9-r)(8-r)(7-r)!} $ $1 = \frac{7}{(9-r)(8-r)} $ $(9-r)(8-r) = 7 $ $72 - 9r - 8r + r^2 = 7 $ $r^2 - 17r + 65 = 0 $ Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $r = \frac{17 \pm \sqrt{289 - 260}}{2} = \frac{17 \pm \sqrt{29}}{2} $ Since $r$ must be a non-negative integer,there is no such $r$ that satisfies the equation. Thus,the correct option is $D$.

If $8 \cdot {}^{7}P_{r} = 7 \cdot {}^{8}P_{r-1}$,what is the value of $r$?

Similar Questions

The number of even numbers greater than $1000000$ that can be formed using all the digits $1, 2, 0, 2, 4, 2, 4$ is

The number of natural numbers lying between $1012$ and $23421$ that can be formed using the digits $2, 3, 4, 5, 6$ (repetition of digits is not allowed) and divisible by $55$ is $....$

The number of positive integral solutions of the equation $xyz = 3000$ is

The sum of all $4$-digit numbers that can be formed using the digits $2, 3, 4, 5, 6$ without repetition is:

$A$ five-digit number divisible by $3$ is to be formed using the digits $0, 1, 2, 3, 4, 5$ without repetition. The total number of ways this can be done is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module