Permutations and Combinations Question (1 Viewer)

Ambility · Nov 30, 2015

In how many ways can the host and hostess and 4 visitors be arranged around a round table if the host and hostess sit next to each other?

My line of thinking:

$\\$Consider host and hostess as one group. There are 5 groups, and within the host and hostess, they can be arranged 2! ways.$\\5! \times 2!\\$The table has 6 chairs, so you can rotate the group six times and get the same result.$\\\frac{5! \times 2!}{6} = 40$

The answer given in the answers section is 48. What did I do wrong?

InteGrand · Nov 30, 2015

Ambility said:
In how many ways can the host and hostess and 4 visitors be arranged around a round table if the host and hostess sit next to each other?

My line of thinking:

$\\$Consider host and hostess as one group. There are 5 groups, and within the host and hostess, they can be arranged 2! ways.$\\5! \times 2!\\$The table has 6 chairs, so you can rotate the group six times and get the same result.$\\\frac{5! \times 2!}{6} = 40$

The answer given in the answers section is 48. What did I do wrong?

$The answer is $2!\cdot 4!=48$; there are $2!$ ways to arrange the couple (either the hostess is on the \textsl{left} of the host, or on the \textsl{right} of the host), and for each of these arrangements of the couple, we can arrange the remaining four people in $4!$ ways.$

InteGrand · Nov 30, 2015

Ambility said:
In how many ways can the host and hostess and 4 visitors be arranged around a round table if the host and hostess sit next to each other?

My line of thinking:

$\\$Consider host and hostess as one group. There are 5 groups, and within the host and hostess, they can be arranged 2! ways.$\\5! \times 2!\\$The table has 6 chairs, so you can rotate the group six times and get the same result.$\\\frac{5! \times 2!}{6} = 40$

The answer given in the answers section is 48. What did I do wrong?

$$The incorrect thing with your reasoning is dividing by 6. Instead, you should have divided by 5, because there are 5 ``groups'' here. It's the number of groups that is relevant, rather than the number of chairs. So even though there's 6 chairs in total, it's as though only 5 chairs are there since there are 5 groups. This is why for placing $n$ people in a circle, we divide by $n$ (i.e. $\frac{n!}{n}=(n-1)!$), because $n$ is the no. of groups being considered (each group in that case is just 1 person). So for your one, it's kind of like placing 5 ``people'' around a table (so divide by 5), it's just that for one of the ``people'', we have $2!$ ways too.$

Ambility · Dec 1, 2015

InteGrand said:
$$The incorrect thing with your reasoning is dividing by 6. Instead, you should have divided by 5, because there are 5 ``groups'' here. It's the number of groups that is relevant, rather than the number of chairs. So even though there's 6 chairs in total, it's as though only 5 chairs are there since there are 5 groups. This is why for placing $n$ people in a circle, we divide by $n$ (i.e. $\frac{n!}{n}=(n-1)!$), because $n$ is the no. of groups being considered (each group in that case is just 1 person). So for your one, it's kind of like placing 5 ``people'' around a table (so divide by 5), it's just that for one of the ``people'', we have $2!$ ways too.$

It still doesn't really make sense in my head as to why we divide by 5 instead of 6. When I imagine these problems I usually think of the amount of arrangements of people in a line and then think of how many ways these arrangements can be represented on a circle. Because each "straight line arrangement" can be rotated n times on a circle, and still have the same arrangement we must divide by n because there are n repetitions of the same arrangement.

I guess the fundamental thing I struggle with is, why do we consider groups instead of the people themselves?

leehuan · Dec 1, 2015

The host and the hostess are arranged as one person, because they must always sit together no matter what.

The 4! occurs as a result of circular arrangement (where one person must be fixed), as opposed to a straight line arrangement which would have the 5! untouched. However, if the host and the hostess DIDN'T sit next to each other then
-> We don't arrange them as one person because they can sit anywhere
-> And there isn't the 2! for who sits on the left and the right.

leehuan · Dec 1, 2015

(Gonna make a new post this time instead of edit an original

When two people sit together, we arrange them together because if we arrange them to be single, we have to somehow eliminate the cases where they are NOT sitting together. For convenience sake, by treating them as one person we don't have to insert a supposed subtraction sign.

Ambility · Dec 1, 2015

leehuan said:
(Gonna make a new post this time instead of edit an original

When two people sit together, we arrange them together because if we arrange them to be single, we have to somehow eliminate the cases where they are NOT sitting together. For convenience sake, by treating them as one person we don't have to insert a supposed subtraction sign.

I get that you treat them as one person and then swap their positions. That's why my original working was $\frac{2!\times 5!}{6}$ instead of $\frac{6!}{6}$ I don't get why it is a five on the denominator instead of six.

leehuan · Dec 1, 2015

Like I said, the people are arranged together. If they're arranged together, you don't have a sixth person. The people you have are the 4 guests, and the one hosthostess

Total amount of people in your circular arrangement = 5

InteGrand · Dec 1, 2015

Ambility said:
I guess the fundamental thing I struggle with is, why do we consider groups instead of the people themselves?

$This is because our method for doing this Q. is first arrange the `groups', and for each group-arrangement, we have $2!$ sub-arrangements for the ordering of the couple. So when we first order the groups, it's $\frac{5!}{5}$ (as there are 5 groups), and for each of these $\frac{5!}{5}$ arrangements, there are $2!$ arrangements for the couple, so a total of $2!\times \frac{5!}{5}$ as the answer.$

Ambility · Dec 1, 2015

InteGrand said:
$This is because our method for doing this Q. is first arrange the `groups', and for each group-arrangement, we have $2!$ sub-arrangements for the ordering of the couple. So when we first order the groups, it's $\frac{5!}{5}$ (as there are 5 groups), and for each of these $\frac{5!}{5}$ arrangements, there are $2!$ arrangements for the couple, so a total of $2!\times \frac{5!}{5}$ as the answer.$

That makes a little bit for sense now, thank you.

Permutations and Combinations Question (1 Viewer)

Ambility

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