Split 6 people into 3 different sized rooms - difference between two solutions

Question

There are 3 rooms in a hostel: a quadruple, a triple and a double one. The hostel is empty when 6 people arrive and fill the rooms randomly. What is the chance that a room stays empty?

I counted it this way: $$P(A) = 1-\frac{{6 \choose 2} \cdot \left({4 \choose 3} + {4 \choose 2} + {4 \choose 1}\right) + {6 \choose 1} \cdot \left({5 \choose 3} + {5 \choose 2} + {5 \choose 1}\right)}{3^6}$$

but my coursebook has a different solution: $$P(A) = \frac{7 \cdot 6! + 6!}{{9 \choose 6}*6!}$$

Why is my answer different?

Answer 1

The only way a room will stay empty is that all the people stay in the quadruple and triple rooms, or all people stay in quadruple and double rooms. So what are the possible combinations to store six people in seven (the sum of beds inside the quadruple and triple rooms) beds? (Case 1)

${{7}\choose{6}}=7$

Indeed what are the possible combinations to store six people in six (the sum of beds inside the quadruple and double rooms) beds? (Case 2)

${{6}\choose{6}}=1$

Now we have to compare those combinations with the total number of combinations in which six people can be stored in nine beds:${9}\choose{6}$

Now simply doing favorable cases over possible cases we get

$\frac{{{7}\choose{6}}+{{6}\choose{6}}}{{9}\choose{6}}=\frac{7+1}{{9}\choose{6}}$

I really don't know why your book is considering the order of possible combinations. If you consider the order you have to add the possible way to order six items: $6!$.

So your book multiplies combinations of case 1, those of case 2 and total combinations for $6!$. If you do that you get:

$\frac{7*6!+6!}{{{9}\choose{6}}*6!}$

but as you can see, you can easily semplify $6!$ in the fraction, the result will be the same.

Answer 2

There are 9 places and 6 people.

The number of possible outcomes: ${9 \choose 6}=84$

If the double room is empty 7 places remain: ${7 \choose 6}=7$ cases. If the triple room is empty 6 places remain: ${6 \choose 6}=1$ case. The quadruple room cannot be empty .

The number of favourable cases: 7+1=8

So the solution in your coursebook is correct.

Answer 3

In your view there are $3^6$ possibilities in total. That is not true. If e.g. $2$ persons have chosen the double then the others cannot go there anymore. Also it is better here to think of the persons as indistinguishable. This in not only in numerator but also in denominator.

If $Q,T,D$ denote the events that respectively quadruple, triple, double stays empty then $Q,T,D$ are mutually exclusive and $Q=\varnothing$ so that: $$\mathsf P(Q\cup T\cup D)=\mathsf P(T)+\mathsf P(D)=\frac{\binom{6}{6}}{\binom{9}{6}}+\frac{\binom{7}{6}}{\binom{9}{6}}=\frac{8}{\binom{9}{6}}$$