In the first step we assign two of the four slots to

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In the first step, we assign two of the four slots to be green. There are ( 4 2 ) = 6 ways of doing this: GG?? ?GG? ??GG G?G? ?G?G G??G As a second step, we take this setup and assign B to one of the blank spots and R to the other. There are P (2 , 2) = 2 ways of doing this. For instance, GG ?? can be turned into either GGBR or GGRB . This again gives us ( 4 2 ) · 2 = 12 possible orderings. Question 3. Now suppose someone is red-green colorblind, and the Emerald, Teal, and Red balls all look indistinguishably Gray. How many possible orderings are there? First approach: We start with the 24 original orderings, and put them in groups that the colorblind person can’t tell apart. For instance, ETBR, TEBR, ERBT, TRBE, REBT, RTBE 1
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are all indistinguishable to this person. There are 3! = 6 ways of permuting the E, T, R balls, so each of these groups contains 6 of the orginal orderings. There- fore there are 24 / 6 = 4 possible orderings: GGGB GGBG GBGG BGGG Second approach: We use a two step process. In the first step, we choose 3 of the 4 positions to contain Gray balls. In the final step, we assign the Blue ball to the remaining spot. There are ( 4 3 ) = 4 ways of doing the first step and only 1 way of doing the second, so there are ( 4 3 ) · 1 = 4 possible orderings.
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