Question

If nP5 : nP3 = 2:1, find the value of n?​

Answer 1

Answer:

n!/(n-5)! divided by (n!/(n-3)!) = 2/1

n!/(n-5)! * (n-3)! / n! = 2

n! cancel

(n-3)! / (n-5)! = 2

Expand (n-3)!

(n-3)(n-4)(n-5)! / (n-5)! = 2

(n-5)! cancel

(n-3)(n-4) = 2

n^2 - 7n + 12 = 2

n^2 - 7n + 10= 2

(n-5)(n-2) = 0

n = 5 or n = 2,

so, we have the condition that n should be greater than or equal to 5  

n=5

Answer 2

The value of n is 5 or 2.

Given:

$^nP_5:^nP_3$=2:1

To find:

The value of n.

Solution:

The given expression is $^nP_5:^nP_3$.

In terms of factorial, it can be written as follows-

n!/(n-5)!:n!/(n-3)!=2:1.

So, we can write-

$\dfrac{n!}{(n-5)!}/\dfrac{n!}{(n-3)!}=2:1\\\dfrac{n!}{(n-5)!}×\dfrac{(n-3)!}{(n)!}=2:1\\\dfrac{(n-3)!}{(n-5)!}=2:1\\\dfrac{(n-3)(n-4)(n-5)!}{(n-5)!}=2:1\\(n-3)(n-4)=2\\n^2-7n+12=2\\n^2-7n+10=0\\n^2-5n-2n+10=0\\n(n-5)-2(n-5)=0\\(n-5)(n-2)=0\\n=5\\n=2$

The value of n is 5 or 2.

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