Hello hello! I am practicing with my textbook\'s example, but I can\'t figureout

Question

Hello hello! I am practicing with my textbook's example, but I can't figureout why what it says is right.. A manager must select 4 employees for promotion. 12 employeesare eligible. a) in how many ways can the 4 be chosen? The textbook says because there is no reason to consider theorder, so we use combinations. But why can't we use permutation? There are 4 empty places for12. For the 1st place, there will be 12 choices, 2nd place 11choices, 3rd place 10 choices and 4th one 9 choices. No??? I can'tget this thing correctly. b) in how many ways can 4 employees be choosen (from 12) to beplaces in 4 different jobs? Here I agree we use permutation. But why is it12P4 and not4P4?? thanks for your help!! Hello hello! I am practicing with my textbook's example, but I can't figureout why what it says is right.. A manager must select 4 employees for promotion. 12 employeesare eligible. a) in how many ways can the 4 be chosen? The textbook says because there is no reason to consider theorder, so we use combinations. But why can't we use permutation? There are 4 empty places for12. For the 1st place, there will be 12 choices, 2nd place 11choices, 3rd place 10 choices and 4th one 9 choices. No??? I can'tget this thing correctly. b) in how many ways can 4 employees be choosen (from 12) to beplaces in 4 different jobs? Here I agree we use permutation. But why is it12P4 and not4P4?? thanks for your help!!

Explanation / Answer

Now, let's do the same experiment, but this time make the prizesthe same (a case of Sunny D). Here are all the outcomes:

AB, AC, BC   ---------> 3 outcomes

Notice that AB is the same as BA (they are both getting Sunny Dso the order is irrelevent), therefore we don't have to list ittwice.   So, here the order of the selection does notmatter, therefore we could solve thisby  3C2 = 3, which agrees with ouranswer that we listed out.

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