Permutation question

[sachin_yadav]

Hi Everyone,

Please help me in the following question. I thought it was an easy one but got me wrong.

In how many ways can N students be seated in a row with N seats ?

(1) |N-6| = 3
(2) N^2 = 7N + 18

Answer is B. My answer was D.

I don't understand why first option is not correct. The following way in which i solved the first option:-

|N-6|=3
N-6=3
N=9

But explanation says:-

|N-6| = 3
N-6=3 Or N-6=-3
N=9 or N=3

[cans]

n students, n seats. Thus ways = n! (we need to find n)
a)|n-6|=3
n-6=3 ->n=9. thus 9! ways
or n-6=-3. n=3, 3! ways.
insufficient
n)n^2 - 7n -18=0
n^2 - 9n +2n-18=0
n(n-9) + 2(n-9)=0
(n+2)(n-9)=0
n = -2 or 9
n is no. of students. thus >0
n=9
9! ways
Sufficient
IMO B

In a) we are given modulus, n=3 and 9 both satisfy the condition.

[galaxian]

Whenever we open a Modulus to solve an equation like:

|N-6| = 3

We proceed as :

N-6 = +/-3.
So a Modulus Equation actually consists of 2 Eqns :
N-6 = +3 &
N-6 = -3.

Hence, 2 solutions.
Rest of the solution, same as Cans's.

[Whitney Garner]

Hi Everyone,

Please help me in the following question. I thought it was an easy one but got me wrong.

In how many ways can N students be seated in a row with N seats ?

(1) |N-6| = 3
(2) N^2 = 7N + 18

Answer is B. My answer was D.

I don't understand why first option is not correct. The following way in which i solved the first option:-

|N-6|=3
N-6=3
N=9

But explanation says:-

|N-6| = 3
N-6=3 Or N-6=-3
N=9 or N=3

We always need to remember that the absolute value has "hidden" the sign for us (in the same way as an even power). So we have to think of the 2 cases:

(1) the stuff inside the absolute value was actually positive (so we can ignore the braces)

N-6 = 3
N = 9

(2) the stuff inside the absolute value was actually negative (so we have to manually remove the negative by multiplying by another negative:

-(N-6) = 3
N - 6 = -3
N = 3


Whit