# Permutations versus combinations, order or unordered (problem submitted).

A tourist wants to visit six of America’s ten largest cities. In how many ways can she do this if the order of her visits is (a) important, (b) not important?

For part (a), I believe the answer is a combination [10 choose 6]. But I'm not sure about part (b).

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Can you tell us (from your instructor or textbook) what exactly is meant by a permutation and a combination? –  David Feb 3 at 3:33
Number of permutations of k objects from a set of size n, or Pn,k for short is n!/(n-k)! For combinations Cn,k is short for the number of combinations of k things taken from a set of n. Cn,k is often written as n!/((k!)(n-k)!). –  MMP Feb 3 at 3:39
Yes but what do permutations and combinations mean? The formulae give you the number of permutations and combinations - this is the easy bit! –  David Feb 3 at 3:42

The word "important" here probably implies that order is important to the reader (for the purposes of counting) rather than the tourist.

For (a), this means we count the possible distinct orderings (permutations) of the destinations, whereas for (b) we count only the possible sets of destinations (combinations) [or we count distinct orderings of the same destinations together].

[PS. If "important" were interpreted to mean important to the tourist (which I don't think is unreasonable), then your proposed answer to (a) would be correct: there are $\binom{10}{6}$ unordered sets of destinations, and the tourist visits them according to her preferences.]
(a) is $10\choose6$, combinations
(b) is ${10\choose 6} \times 6!$ because there are $6!$ orders to visit the cities