Evaluate the value of $n$ in $\, ^{n+2} P_5=18 \times \, ^{n}P_4$

Question

Suppose $\, ^{n+2} P_5=18 \times \, ^{n}P_4$. How can we evaluate n?

I tried expanding it like this:

$$(n+2)(n+1)(n+0)(n-1)(n-2)=18 \cdot n(n-1)(n-2)(n-3)$$

Then I try multiplying all of them above, but it gets longer....and now um confused.

I thoyght of dividing the R.H.S by the L.H.S (leaving the $18$ in the R.H.S) so that I could cancel some of them out.

Not sure about it.

Answer 1

Matthew Conroy's suggestion in the comment is right on the nail. Often we have some polynomial and we go to great effort to factor it. Here we have polynomials already factored, so we should leave them alone! We have:

$$(n+2)(n+1)(n+0)(n-1)(n-2)=18 \cdot n(n-1)(n-2)(n-3)$$

and we can immediately cancel the factors of $n, n-1,$ and $n-2$ from both sides, observing along the way that $n=0,1,2$ are all trivial solutions. The cancellation leaves us with:

$$(n+2)(n+1)=18 \cdot (n-3)$$

and now we should expand the polynomials and collect terms, obtaining $$n^2-15n +56 = 0$$

which we can solve by the usual methods to obtain the solutions $n=7,8$.

Answer 2

First, are there restrictions on $n$? For example, n=0,1,2 could lead to each side evaluating to zero though one could argue the 4 on the right hand side implies $n\geq4$.

$(n+2)(n+1)=18(n-3)$ is the reduced equation.

Noticing the 18 on the R.H.S. gives me the idea of trying n=7 or 8 as the first couple of ideas:

L.H.S.: $9*8 = 72 $

R.H.S.: $18*4 = 72$

Trying n=8 gives:

L.H.S.: $10*9=90$

R.H.S.: $18*5=90$

For higher n like 16,17 the terms are too different to work out. 18*17 compared to 18*13 isn't going to be equal, nor would 19*18 compared to 18*14 as these would have a factor in common here.

Thus, $n=0,1,2,$ or $7,8$ would be solutions assuming one doesn't properly put a lower bound on n.

Answer 3

We have $P(n+2,5) = 18 \cdot P(n,4)$, I believe. Note that $P(x,k)=\frac{x!}{(x-k)!}$ for $x,k \in \mathbb{N}$ such that $x>k$, so that this equation can be rewritten as $$\frac{(n+2)!}{((n+2)-5)!}=\frac{(n+2)!}{(n-3)!}=(n+2)(n+1) \cdot n(n-1)(n-2) \\= 18 \cdot \frac{n!}{(n-4)!}=18 \cdot n(n-1)(n-2)(n-3).$$ Note from here that $n=0,1,2$ are trivial solutions (substituting these result in $0=0$). Thus, we have (for $n \neq 0,1,2,3$) $$\frac{(n+2)(n+1)}{n-3}=18.$$ This should give you a good start. Rearrange this and collect like terms to obtain a polynomial equation that you can solve for $n$.

Answer 4

Note that some terms on the left and right side of the equation will cancel.

$$\require{cancel} (n+2)(n+1)\cancel{(n+0)}\cancel{(n-1)}\cancel{(n-2)}=18 \cdot \cancel{n}\cancel{(n-1)}\cancel{(n-2)}(n-3)$$

You'll get a quadratic in $n$:

$$n^2 + 3n + 2 - 18n + 54 = n^2 - 15n + 56 = (n-7)(n-8) = 0$$