Question 4 we now have 10 balls 3 are green 2 are red

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Question 4. We now have 10 balls. 3 are Green, 2 are Red, and the remaining 5 are Yellow, Blue, Purple, Mauve, and Chartreuse (and all 5 can be distinguished from each other). How many possible orderings are there? First approach: If the Green and Red were distinguishable, there would be 10! possible orderings. We can then group these into indistinguishable collec- tions. For instance one ordering is G 1 G 2 Y BR 1 MR 2 PG 3 C. Then all rearrangements of the Green and Red balls are indistinguishable: G 1 G 2 Y BR 1 MR 2 PG 3 C G 2 G 3 Y BR 1 MR 2 PG 1 C G 3 G 1 Y BR 1 MR 2 PG 2 C G 3 G 2 Y BR 1 MR 2 PG 1 C G 2 G 1 Y BR 1 MR 2 PG 3 C G 1 G 3 Y BR 1 MR 2 PG 2 C G 1 G 2 Y BR 2 MR 1 PG 3 C G 2 G 3 Y BR 2 MR 1 PG 1 C G 3 G 1 Y BR 2 MR 1 PG 2 C G 3 G 2 Y BR 2 MR 1 PG 1 C G 2 G 1 Y BR 2 MR 1 PG 3 C G 1 G 3 Y BR 2 MR 1 PG 2 C Given one of the 10! original colorings, there are 3! = 6 ways of arranging the Green balls and 2! = 2 ways of arranging the Red balls, for a total of 12 possible orderings. So there are 10! 3!2! orderings where we can’t distinguish the green balls from each other or the blue balls from each other.
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