If {}^n{P_5} = 20 cdot {}^n{P_3},then n = | vedclass

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

Question

(B) Given the equation: ${}^n{P_5} = 20 \cdot {}^n{P_3}$Using the formula ${}^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 20 \cdot \frac{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given the equation: ${}^n{P_5} = 20 \cdot {}^n{P_3}$Using the formula ${}^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}$Dividing both sides by $n!$ (since $n! 
eq 0$):$\frac{1}{(n-5)!} = \frac{20}{(n-3)!}$$(n-3)! = 20 \cdot (n-5)!$$(n-3)(n-4)(n-5)! = 20 \cdot (n-5)!$Since $(n-5)! 
eq 0$,we can cancel it from both sides:$(n-3)(n-4) = 20$$n^2 - 7n + 12 = 20$$n^2 - 7n - 8 = 0$$(n-8)(n+1) = 0$This gives $n = 8$ or $n = -1$.Since $n$ must be a positive integer and $n \ge 5$ for ${}^n{P_5}$ to be defined,we discard $n = -1$.Therefore,$n = 8$.

If ${}^n{P_5} = 20 \cdot {}^n{P_3}$,then $n = $

Similar Questions

Eleven books consisting of $5$ Mathematics,$4$ Physics,and $2$ Chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books of the same subject are all together is

How many five-digit numbers can be formed using the digits $2, 0, 4, 3, 8$ if repetition of digits is not allowed?

If a polygon has $44$ diagonals,then the number of its sides are

The value of $r$ for which $^{20}C_r \cdot ^{20}C_0 + ^{20}C_{r-1} \cdot ^{20}C_1 + ^{20}C_{r-2} \cdot ^{20}C_2 + \dots + ^{20}C_0 \cdot ^{20}C_r$ is maximum is:

The number of arrangements of the letters of the word $SATAYPAUL$ such that no two $A$ are together and the middle letter is a consonant,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module