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# Advanced Math/r-permutations and r-combinations of grouped objects

Question
1. How is the number of r-permutations of n objects including n1,n2,n3,… identical objects calculated? (r<n)

For example, what is the number of ways in which 3 yellow balls, 2 red balls and 4 green balls can be put in 5 baskets, each in one basket, using five of the balls. (in this example, n=9 , n1=3 , n2=2 , n3=4 and r=5.)

2. How is the number of r-combinations of such objects calculated? (where order does not matter at all, even between different groups)

Thank you very much in advance.

Think of 1 as arranging things in order with the borders between the baskets as an object.
Now one basket border has to go at either end.  For 5 baskets side by side, there is one border between each of the baskets and one at either end.  Since there are 5 baskets, there are 5-1=4 borders between that can be thought of as being arranged with the balls.

The number of objects to arrange is then 3 yellow + 2 red + 4 green + 4 borders = 13 objects.  Since there is no particular way that order is involved, that is 13!/(3!2!4!4!).

Cancelling one of the 4! terms makes it 13*12*11*10*9*8*7*6*5/(3*2*1*2*1*4*3*2*1).
The 1st 3*2 on bottom cancels a 6 on top, next are 2 and 4, which cancel with an 8 on top,
the 1's in the denominator disappear, the last two terms in the denominator are 2 and 3.
Note that 12/(2*3) = 2.  We cancelled a 6, an 8, and made the 12 a 2 while cancelling all of the terms in the denominator.  That leaves 13*2*11*10*9*7*5.  Now if you know that 7*11*13 = 1001, that leaves 2*10*9*5*1001.  Since 2*5 = 19, that is 900*1001, which is 900,900.

That is the number of combinations.  A permutation is an order combination.
Thus, we would think of each of the 9 balls as a separate ball.  Then remember that each of the partitions could go anywhere, which makes for 13 items.  Thus, it would be 13!/4! since
13 - 9 = 4.  That works out to  be 259,459,200.  That is 288 times as many for each un-ordered combination.

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#### Scott A Wilson

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