If frac{n!}{2!(n-2)!} and frac{n!}{4!(n-4)!} | vedclass

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

Question

(B) Given the ratio: $\frac{n!}{2!(n-2)!} : \frac{n!}{4!(n-4)!} = 2:1$.This can be written as: $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!} = \frac

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given the ratio: $\frac{n!}{2!(n-2)!} : \frac{n!}{4!(n-4)!} = 2:1$.This can be written as: $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!} = \frac{2}{1}$.Canceling $n!$ from the numerator and denominator: $\frac{4!(n-4)!}{2!(n-2)!} = 2$.Expanding the factorials: $\frac{4 	imes 3 	imes 2!}{2!} 	imes \frac{(n-4)!}{(n-2)(n-3)(n-4)!} = 2$.Simplifying the expression: $\frac{12}{(n-2)(n-3)} = 2$.Dividing both sides by $2$: $\frac{6}{(n-2)(n-3)} = 1$.So,$(n-2)(n-3) = 6$.Expanding the quadratic equation: $n^2 - 5n + 6 = 6$.$n^2 - 5n = 0$.$n(n-5) = 0$.Since $n$ must be at least $4$ for the term $(n-4)!$ to be defined,we have $n = 5$.

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$,find the value of $n$.

Similar Questions

How many different necklaces can be formed with $6$ white and $5$ red beads?

There are seven candidates for $3$ posts. In how many ways can the posts be filled?

In how many ways can the letters of the word '$UNIVERSAL$' be arranged? In how many of these will $E, R, S$ always occur together?

In how many different ways can the letters of the word '$THERAPY$' be arranged so that the vowels never come together?

The number of arrangements of the letters of the word $PALANHAR$ in which no two vowels are together and exactly two vowels are at odd places,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module