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

Assuming that no two consecutive digits are same,the number of $n$-digit numbers is:

The number of words that can be formed by using the letters of the word $CALCULATE$ such that each word starts and ends with a consonant,are

If all the possible $3$-digit numbers are formed using the digits $1, 3, 5, 7, 9$ without repeating any digit,then the number of such $3$-digit numbers which are divisible by $3$ is

How many words of $7$ letters can be formed by selecting $4$ consonants and $3$ vowels from $6$ consonants and $5$ vowels?

The number of numbers,strictly between $5000$ and $10000$,that can be formed using the digits $1, 3, 5, 7, 9$ without repetition is $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module