Question

Three gnomes and three elves sit down in a row of six chairs. If no gnomes will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?
A.$54$
B.$72$
C.$80$
D.$60$

Answer 1

We can sit them in alternate seats so that no gnome will sit next to another gnome and no elf will sit next to another elf. So the arrangement will be (GEGEGE) or (EGEGEG). Now we know that- to select r number of things out of n total number of things, we use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$
The formula of combination is given as-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$Where n=total number of things and r=no. of things to be selected. Use these two formulas to find the number of ways the gnomes and elves can sit together.

Complete step-by-step answer:
Given, there are a total of $6$persons in which three are gnomes and three are elves and they sit down in a row of six chairs. No gnomes want to sit next to another gnome and no elf wants to sit next to another elf. We have to find the number of ways the elves and gnomes can sit.
Since no gnomes or elves are to sit together, they will sit alternately to each other. So the arrangement will be as follows- (GEGEGE) or (EGEGEG) where G denotes gnome and E denotes Elf.
Suppose that the first gnome sits in a chair so he can sit in any of the six chairs. We know that- to select r number of things out of n total number of things, we use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$
So we get-
$\Rightarrow$ The total number of ways the first gnome can sit is=${}^6{{\text{C}}_1}$
Now, we know that formula of combination is given as-
$\Rightarrow$ ${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$Where n=total number of things and r=no. of things to be selected
So on applying the formula, we get-
$\Rightarrow$ The total number of ways the first gnome can sit is=$\dfrac{{6!}}{{1!5!}}$
We know $n! = n \times \left( {n - 1} \right)! \times ...3,2,1$
So on solving, we get-
$\Rightarrow$ The total number of ways the first gnome can sit is=$\dfrac{{6 \times 5!}}{{1!5!}} = 6$-- (i)
Now, if the first gnome sat on an even-numbered chair then the second gnome has to sit on either of the even-numbered chairs left –(G _ _ _ _ ) So either he will sit on the $4{\text{th}}$ or in $6{\text{th}}$ chair. So the number of ways he can choose his seat=${}^{\text{2}}{{\text{C}}_{\text{1}}}$
On solving, we get-
$\Rightarrow$ The number of ways the second gnome can sit=$\dfrac{{2!}}{{1!1!}} = 2$ --- (ii)
Now, the third gnome will have only one choice-( G _ G _ _)
So the number of ways the third gnome can sit is=$1$ -- (iii)
Now, the first elf will have three choices and he has to choose one seat so he can sit in the total number of ways=${}^{\text{3}}{{\text{C}}_{\text{1}}}$
On solving, we get-
The number of ways the first elf can sit=$\dfrac{{3!}}{{1!2!}} = 3$ -- (iv)
Now, the second elf has only two choices- (G E G _ G _)
So the number of ways the second elf can sit=${}^{\text{2}}{{\text{C}}_{\text{1}}}$
On solving we get-
$\Rightarrow$ The number of ways the second elf can sit=$\dfrac{{2!}}{{1!1!}} = 2$-- (v)
And now the last elf has only one choice so he can sit in the number of ways=$1$ -- (v)
On multiplying the terms of eq. (i), (ii), (iii), (iv) and (v), we get-
$\Rightarrow$ The total number of ways the gnomes and elves can sit=$6 \times 2 \times 1 \times 3 \times 2 \times 1$
On multiplication, we get-
$\Rightarrow$ The total number of ways the gnomes and elves can sit=$72$
The gnomes and elves can sit together in six chairs in $72$ ways.

Note: We can also solve this in the following ways-
Since the arrangement is(GEGEGE) or (EGEGEG) so the gnomes can sit in three places and they can also interchange their seats between themselves in $3!$ ways so the total number of ways the gnomes can sit is $3! \times 3!$
Now, the elves will sit in the same way so the total number of ways both elves and gnomes can sit=$2 \times 3! \times 3!$ =$72$