If 7 cdot ^nP_3 = 20 cdot ^{n+1}P_2,then n = | vedclass

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

Question

(D) Given the equation: $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we get:$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Given the equation: $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we get:$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot \frac{(n+1)!}{(n-1)!}$$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot \frac{(n+1)n!}{(n-1)(n-2)(n-3)!}$Dividing both sides by $n!$ and multiplying by $(n-3)!$:$7 = \frac{20(n+1)}{(n-1)(n-2)}$$7(n^2 - 3n + 2) = 20n + 20$$7n^2 - 21n + 14 = 20n + 20$$7n^2 - 41n - 6 = 0$Factoring the quadratic equation:$7n^2 - 42n + n - 6 = 0$$7n(n - 6) + 1(n - 6) = 0$$(7n + 1)(n - 6) = 0$Since $n \in \mathbb{N}$,we have $n = 6$.

If $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$,then $n = \dots$

Similar Questions

In how many ways can $5$ red,$4$ blue,and $1$ green ball be arranged in a row?

How many $3$-digit numbers can be formed from the digits $1, 2, 3, 4,$ and $5$ assuming that repetition of the digits is not allowed?

Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom-most objects?

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

If all possible numbers are formed by using the digits $1, 2, 3, 5, 7$ without repetition and they are arranged in descending order,then the rank of the number $327$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module