Sheldon Ross - Probability (2nd chapter)

Question

Poker dice is played by simultaneously rolling 5 dice.Show that

(a) P{no two alike} = .0926;
(b) P{one pair} = .4630;
(c) P{two pair} = .2315;
(d) P{three alike} = .1543;
(e) P{full house} = .0386;
(f) P{four alike} = .0193;
(g) P{five alike} = .0008.

My doubt is how is this different from poker played by cards? Is it that when we play via dice the order of the dice outcome matters which doesn't in case of cards.(i.e the order of cards in hand). So if we think the case of having one pair as

<6 6 5 1 2> and  <6 5 1 2 6> both outcomes are different?

but <6(1st six) 5 1 2 6 (second six)> and <6(2nd six) 5 1 2 6(1st six)> both are same,since its the same one pair but the pairs doesn't generate new outcome when position is changed.

i.e for question (b) probability is  = 6C1 (for selecting pair outcome) * 5C2 (possibility of selecting a pair of hand out of five,but order does not matter when pair partners are exchanged) * 5 * 4* 3 (because order matters of the outcomes of other non-pairs) this total divided by 6^5. Is my assumption true?

Comments

I think all 5 dices are thrown at the same time so (6,1,2,3,6) is same as (6,6,1,2,3) and to it all permutations.

If it is explicitly mentioned or can be understood that 1 die thrown 5 time than order matters.