Find r if ^{5}P_{r} = ^{6}P_{r-1}. | vedclass

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

Question

(C) Given: $^{5}P_{r} = ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$$\frac{5!}{(5

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given: $^{5}P_{r} = ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$$\frac{5!}{(5-r)!} = \frac{6 	imes 5!}{(7-r)!}$$\frac{1}{(5-r)!} = \frac{6}{(7-r)(6-r)(5-r)!}$$1 = \frac{6}{(7-r)(6-r)}$$(7-r)(6-r) = 6$$42 - 7r - 6r + r^{2} = 6$$r^{2} - 13r + 36 = 0$$(r-4)(r-9) = 0$$r = 4$ or $r = 9$Since $^{n}P_{r}$ is defined for $0 \leq r \leq n$,for $^{5}P_{r}$,we must have $r \leq 5$.Therefore,$r = 9$ is rejected.Hence,$r = 4$.

Find $r$ if $^{5}P_{r} = ^{6}P_{r-1}$.

Similar Questions

The number of five-digit numbers,greater than $40000$ and divisible by $5$,which can be formed using the digits $0, 1, 3, 5, 7,$ and $9$ without repetition,is equal to:

How many words can be formed with the letters of the word $MATHEMATICS$ by rearranging them?

How many numbers can be formed using the digits $0, 1, 2, 3, 4, 5$ that are greater than $3000$ (repetition is not allowed)?

How many $4$-letter codes can be formed using the first $10$ letters of the English alphabet,if no letter can be repeated?

The number of ways in which $n$ boys and $n$ girls can be arranged in a row such that all the boys are together and all the girls are also together is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module