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There are 10 skiers that want to go up the mountain but there are only 4 chairs that hold 2 people each. How many different combinations can I make? I'm currently only asking if the order they sit in the chairs and the order that the chairs go up does NOT matter (no permutations).

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    I am asking how to group 10 objects into duos for 4 available positions and it has been answered satisfactorily. – 2017-01-21
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    Other analogies for the question: You have 10 employees for your theme park. The park has 4 rides and each ride must have 2 employees manning them. How many different schedules could you produce? You have ten food ingredients (single serving) and need to make a menu with 4 dishes that each contain 2 of the ingredients each. How many different menus could you make? – 2017-01-21

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From what you say, it seems that the chairs are to be treated as identical, in which case:

Choose and permute $8$ people,
then remove permutations of the $4$ groups, and the duos in each chair, thus

$\dfrac{\binom{10}{8}8!}{4!(2!)^4}\;$ or more succinctly,$\;\;\dfrac{^{10}P_8}{4!(2!)^4} = 4725$

Alternate formulation:

$\dfrac{\binom{10}2\binom82\binom62\binom42}{4!}$,

or more succinctly, using the multinomial formula, $\dfrac{\binom{10}{2,2,2,2,2}}{4!}$

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    +1 for the multiple formulations! For those of us without the bigger picture yet, can you break down which part of the first formulation is removing permutations of 4 groups and which part is removing the duos in each chair? I'm hoping one day to be able to just see it! Thanks! – 2017-01-20
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    $4!$ removes the permutations of the four groups, and $2!2!2!2! = (2!)^4$ removes the permutations of the duos in each chair. In the second formula, since we are forming the groups using combinations, only the permutations of the four groups need to be removed. – 2017-01-21
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    I have also added a succinct version of the formula for the first approach to clearly show that we are using permutations for counting for the numerator. – 2017-01-21