1
$\begingroup$

In how many ways can 4 people occupy 10 chairs in a row if no two sit on adjacent chairs

I can think of 2 ways to solve this question both giving different answers but I don't know why they're not one-on-one:

Way 1:
4 People take 4 chairs and sit with 3 chairs in the space between them:
_ |P| C_ |P| C_ |P| C_ |P|_ where '_' indicates a space for another chair. Now we have 3 chairs remaining which can be placed in $\binom{5}{3}$ ways. But the people can exchange their seats in 4! ways. Hence total permutations= 10*4!=240

Way 2:
The other way is to keep 6 chairs. Now 4 people can be seated in the 7 spaces between thee chairs. _C_C_C_C_C_C_. There are 4 people and 7 spaces so total permutations=$^7P_4$=840

I want to know why these ways do not lead to the same answer and what arrangements am I missing in the first way.

  • 0
    Presumably the four people are distinct as individuals, so there will be a factor of $4!$ in both answers. – 2017-02-25
  • 0
    @hardmath the 2nd case is anyways a permutation so 4! need not be multiplied separately – 2017-02-25
  • 2
    Your way1 does not count arrangements with *many* chairs next to one another, for example it doesn't count the arrangement |P|CCCC|P|C|P|C|P|, instead only allowing a single "*extra*" chair to be placed in each empty spot. Use stars and bars instead for how to distribute the extra chairs. – 2017-02-25
  • 0
    @JMoravitz Ah...got it, you can post this as an answer – 2017-02-25

2 Answers 2

2

Your way1 does not count arrangements with many chairs next to one another, for example it doesn't count the arrangement |P|CCCC|P|C|P|C|P|, instead only allowing a single "extra" chair to be placed in each empty spot. Use stars and bars instead for how to distribute the extra chairs.

Approaching the same, we have $\underline{~}PC\underline{~}PC\underline{~}PC\underline{~}P\underline{~}$ and we wish to distribute three extra chairs to these five empty spaces (with possibly multiple going to the same empty spot).

By stars and bars we have $\binom{5+3-1}{5-1}=\binom{7}{4}$ number of ways to do so. Multiplying by the number of ways to arrange the persons themselves among the seats labeled $P$ we have a total of $4!\binom{7}{4}=~^7P_4=840$ arrangements, same as the other answer.

1

You have 10 chairs, 4 individuals. They can't sit next to each other. Assuming the chairs aren't in a circular shape, you need to consider dots separating the people which you can arrange in 10 chairs. Since these empty spots can be repeated, as in 2 chairs or more between individuals, this is an unordered with repeat combinatorix question. Furthermore, all must be seated, or 5 separators to accomodate the dots.

This is then (9+5-1)C5 or 13C5 ways to place 4 people on 10 chairs with none next to each other.

Try the above reasoning with 4 chairs and 2 persons which is (3+2-1)C2 which is 6