Question:
Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?
(note: 'C' represents combination, I am new to this so I am not quite efficient with the mathjax formatting)
My Answer: I am quite surprised because my answer seems improbable and I cannot find flaw in the logic at the same time!.. I thought about the problem in this way→
Since, the 6 tourists are to be grouped in 2 groups, one for each guide, this problem reduces to nothing but placing 6 (distinguishable) balls in 2 boxes with the condition that no box may remain empty.. now, obviously that can be done in $*5C1*$ or 5 ways.. and for each way, we can arrange the balls (since tourists are obviously distinguishable) in $6!$ ways.. by this, we cover all the possible cases namely–
$ (1+5)$
$(2+4)$
$(3+3)$
$ (4+2)$
$ (5+1) $
And all possible arrangements of the tourists (for each case) among the 2 guides..
So we can conclude that the answer must be
$5 * 6! $ which is $3600$, which is a huge number for this!
I know, there are other ways to solve this.. but my question is where did I possibly go wrong??
Thank you for your help in advance. 🙂
Best Answer
There are $\frac {6!}{ 1! \times 5! } = 6$ permutations for each of the (1 + 5) groups.
There are $\frac {6!}{ 2! \times 4! } = 15$ permutations for each of the (2 + 4) groups.
There are $\frac {6!}{ 3! \times 3! } = 20$ permutations for the (3 + 3) groups.
Total is $ (6 \times 2) + (15 \times 2) + 20 = 62$.
Which, as another answer says, is $2^6 -2$.