# Thread: No. of ways of arrangement

1. ## No. of ways of arrangement

3 boys, 2 girls and a puppy sit at a round table. In how many ways can they be arranged if the puppy is to be seated between any 2 boys?

Case 1: The girls sit together
No. of ways=(4-1)!(2!)=12

Case2: The girls are seperated
No. of ways=(5-1)!=24

Total no. of ways=12+24=36

2. ## Re: No. of ways of arrangement

Originally Posted by Punch
3 boys, 2 girls and a puppy sit at a round table. In how many ways can they be arranged if the puppy is to be seated between any 2 boys?

Case 1: The girls sit together
No. of ways=(4-1)!(2!)=12

Case2: The girls are seperated
No. of ways=(5-1)!=24

Total no. of ways=12+24=36
If sits are not numerated then you have simple equality :

$N_o=2 \cdot 3 \cdot 3 !=36$

3. ## Re: No. of ways of arrangement

Can i have reasoning of your workings? It looks simple.

4. ## Re: No. of ways of arrangement

Originally Posted by Punch
Can i have reasoning of your workings? It looks simple.
Since there are $3$ boys and $2$ seats the number of arrangements is : $\frac {3 !}{(3-2)! \cdot 2!}=3$ , if you don't take in consideration who is on the left hand side from the puppy and who is on the right hand side from the puppy .

So , to get total number of ways you have to double number : $\frac {3 !}{(3-2)! \cdot 2!}$

Now,in each arrangement there are $3$ more free seats and $1$ boy and $2$ girls...so you have $3 !$ number of ways in which they can sit .

Finally , to get total number of arrangements you have to multiply $2 \cdot \frac {3 !}{(3-2)! \cdot 2!}$ by $3 !$

5. ## Re: No. of ways of arrangement

Hello, Punch!

3 boys, 2 girls and a puppy sit at a round table.
In how many ways can they be arranged
if the puppy is to be seated between any 2 boys?

Assume there are six chairs around the table.

The puppy can sit anywhere (it doesn't matter).
There are 3 choices for the boy on its left.
There are 2 choices for the boy on its right.
The other 3 people (1 boy, 2 girls) can be seated in 3! ways.

Therefore: . $3\cdot2\cdot3! \,=\,36$ ways.