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

It is required that each one of the tourists must choose either of the two guides. This means, therefore, the possible number of groupings is: [tex]2^{6} [/tex]64.

However, if we subtract the 2 people (recall that the guides decided to split) the answer becomes calculable as follows:

You'd get [tex]\frac{6!}{1! * 5!} [/tex] = 6 permutations for each of the groups containing (1+5)
You'd also get [tex]\frac{6!}{2! * 4!} [/tex] = 15 permutations for each group containing (2+4)
then you'd also get [tex]\frac{6!}{3! * 3!} [/tex] = 20 permutations for each group containing (3+3)

Totaling each scenario, we can say:

(6 x2) + (15x2) + 20 = 62

Another way to express this is

[tex]2^{6} [/tex] - 2= 62

What is permutation?

This refers to the various possible ways in which a given amount or number of items, or persons, may be arranged.

See more questions related to permutations in the link below:

https://brainly.com/question/9253208