Find r if ^{5}P_{r} = 2 times ^{6}P_{r-1}. | vedclass

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

Question

(A) Given equation: $^{5}P_{r} = 2 	imes ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = 2 	imes \frac{6!

Vedclass · Accepted Answer

$r = 3$

Vedclass · Answer

(A) Given equation: $^{5}P_{r} = 2 	imes ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = 2 	imes \frac{6!}{(6-(r-1))!}$$\frac{5!}{(5-r)!} = \frac{2 	imes 6!}{(7-r)!}$$\frac{5!}{(5-r)!} = \frac{2 	imes 6 	imes 5!}{(7-r)(6-r)(5-r)!}$Dividing both sides by $5!$ and multiplying by $(5-r)!$:$1 = \frac{12}{(7-r)(6-r)}$$(7-r)(6-r) = 12$$42 - 7r - 6r + r^{2} = 12$$r^{2} - 13r + 30 = 0$$(r-10)(r-3) = 0$So,$r = 10$ or $r = 3$.Since the condition for $^{n}P_{r}$ is $0 \leq r \leq n$,for $^{5}P_{r}$,we must have $r \leq 5$.Therefore,$r = 3$ is the only valid solution.

Find $r$ if $^{5}P_{r} = 2 \times ^{6}P_{r-1}$.

Similar Questions

There are $6$ letters and $3$ post boxes. In how many ways can these letters be posted in the post boxes?

We are to form different words with the letters of the word $INTEGER$. Let $m_1$ be the number of words in which $I$ and $N$ are never together and $m_2$ be the number of words which begin with $I$ and end with $R$,then $m_1/m_2$ is equal to

The number of words that can be written using all the letters of the word "$IRRATIONAL$" is

The number of arrangements of the letters of the word $INDEPENDENCE$ in which all the vowels always occur together is

If the letters of the word $RACHIT$ are arranged in all possible ways and these words are listed in dictionary order,what is the rank of the word $RACHIT$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module