# Thread: Need help solving for 'n': P (n,5) = 42 x P(,3)

1. ## Need help solving for 'n': P (n,5) = 42 x P(,3)

Solve for n
P (n,5) = 42 x P(,3)

.. This is from a grade 12 Data Management course that I am taking online! Other people have tried to help me on a different forum, but I still don't know what to do!

2. Originally Posted by cnmath16
Solve for n
P (n,5) = 42 x P(,3)

.. This is from a grade 12 Data Management course that I am taking online! Other people have tried to help me on a different forum, but I still don't know what to do!
I assume by P(,3) you meant P(n, 3). I assume that the notation represents permutations. You should know a basic formula which allows the equation to be written as

$\displaystyle \frac{n!}{(n-5)!} = 42 \cdot \frac{n!}{(n-3)!}$

$\displaystyle \Rightarrow \frac{1}{(n-5)!} = 42 \cdot \frac{1}{(n-3)!}$

$\displaystyle \Rightarrow (n-3)! = 42 (n-5)!$

Now you have three choices:

Option 1: Guess and check. When you get to n = 10 you'll have the answer.

Option 2: By definition of the factorial, the equation can be written as

$\displaystyle (n-3)(n-4)(n-5)! = 42 (n-5)!$

$\displaystyle \Rightarrow (n-3)(n-4) = 42$

$\displaystyle \Rightarrow n^2 - 7n + 12 = 42 \Rightarrow n^2 - 7n - 30 = 0 \Rightarrow (n-10)(n+3) = 0$.

Therefore n = 10 (the negative solution is rejected).

Option 3: Note that $\displaystyle 42 = 7 \cdot 6$. Therefore $\displaystyle (n-3)! = 7 \cdot 6 \cdot (n-5)!$. This suggests that the left and right hand sides are equal to 7!. Hence n = 10. Check it ..... Correct.