Author Topic: Permutations and combinations (Read 2428 times) Tweet Share

Elisejw · « **on:** February 01, 2018, 06:20:29 pm »

Hello! I'm stuck on this question. Could someone please help me? Thank you!

The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time is 14:5. Find the value of n.

RuiAce · « **Reply #1 on:** February 01, 2018, 06:40:45 pm »

Previously addressed here.

Elisejw · « **Reply #2 on:** February 01, 2018, 06:44:54 pm »

Hello. Sorry, for some reason, I can't see your response. Could you try posting it again?

RuiAce · « **Reply #3 on:** February 02, 2018, 02:44:22 pm »

The link works fine for me. Please show me what happens when you click it

jamonwindeyer · « **Reply #4 on:** February 02, 2018, 08:21:21 pm »

I'll copy and paste the answer here for you Elise

$\text{Backtracking from the given answer,}\\ \text{it would appear as though a straight line arrangement was necessary.}$
If this is the old Fitzpatrick textbook then, well trust it to omit important details.
$\text{There are two ways of starting the question,}\\ \text{but they coincide at the end.}$
$\text{Method 1: This is the standard scenario of picking }n\text{ objects}\\ \text{out of }2n+2\text{, to arrange in a line.}$
An example would be arranging 5 people in a line out of 5.
$\text{By formula, this is just }^{2n+2}P_n.$
$\text{Similarly, if we only have }n\text{ people,}\\ \text{then we have }^{2n}P_n\text{ arrangements instead.}$

Method 2

$\text{Alternatively, we first choose }\textbf{as an unordered selection}\\ n\text{ people out of }2n+2: \binom{2n+2}{2}\text{ arrangements.}$
$\text{After choosing them without any ordering, we assign the ordering manually}\\ \text{in the same way we arrange }n\text{ objects in a straight line.}\\ \text{This can be done in }n!\text{ ways.}$
$\text{So when we deal with }n\text{ out of }2n+2\text{ people, we have }\binom{2n+2}{n} n!\text{ possible outcomes.}$
$\text{Similarly, when we do this with }n\text{ out of only }2n\text{ people},\\ \text{we have }\binom{2n}{n}n!\text{ possible outcomes.}$
Note: $ \binom{N}{K} = \frac{N!}{K!(N-K)!} $

$\text{Using method 1, and noting that }^NP_K = \frac{n!}{(n-k)!},\\ \text{we essentially wish to solve }\frac{^{2n+2}P_n}{^{2n}P_n}=\frac{14}{5}$
$\begin{align*}\frac{^{2n+2}P_n}{^{2n}P_n}&=\frac{14}{5}\\ 5 ^{2n+2}P_n &= 14 ^{2n}P_n\\ 5 \times \frac{(2n+2)!}{(n+2)!} &= 14 \times \frac{(2n)!}{n!}\end{align*}$
$\text{Now, observe that}\\ \begin{align*}\frac{(2n)!}{n!}&= \frac{2n (2n-1)(2n-2)\dots (n+1)(n)(n-1)\dots (2)(1)}{n(n-1)\dots(2)(1)}\\ &= 2n(2n-1)\dots(n+1)\end{align*}\\ \text{and similarly,}\\ \frac{(2n+2)!}{(n+2)!}= (2n+2)(2n+1)\dots (n+3)$
$\text{So our equation becomes}\\ 5\times (2n+2)\dots (n+3) = 14\times (2n)\dots(n+1)\\ \text{Cancelling out a lot of terms reduces this down to}\\ 5(2n+2)(2n+1) = 14(n+2)(n+1)$
$\text{This is now just a quadratic equation.}\\ \begin{align*}20n^2+30n+10&= 14n^2+42n+28\\ 6n^2-12n-18&=0\\ 6(n+1)(n-3)&=0\\ n=-1, 3\end{align*}$
$\text{Since }n\ge 0,\text{ our solution is }n=3.$

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Author Topic: Permutations and combinations (Read 2428 times) Tweet Share

Elisejw

Permutations and combinations

RuiAce

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RuiAce

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jamonwindeyer

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