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 5 partitions (Posted on 2012-11-21)
Grandma Jones bought 15 identical chocolate bars for her six grandchildren (named A,B,C,D,E and F).

How many possible ways to divide those bars are there for each of the following exclusive cases:

a) Each of the grandchildren gets a distinct number of bars
b) Only two of the grandchildren get the same number of bars
c) Each of the grandchildren gets at least 1 chocolate bar
d) No restrictions
e) A gets more than B, no other restrictions

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 By hand | Comment 1 of 3
**I'll be very surprised if there are no errors here but I was careful!**

I made a list and found 83 unordered arrangements of 5 non-negative integers whose sum is 15.  My list goes from (15,0,0,0,0) to (3,3,3,3,3).
Each of these can be ordered in a number of ways depending on the pattern.

Part a)  There are 7 unordered.
(9,3,2,1,0) ... (5,4,3,2,1)
each has 5!=120 permutations.  7*120 = 840

Part b)  There are 37 unordered.
(12,2,1,0,0) ... ( 5,4,3,3,0)
each has 5!/2! = 60 permutations.  37*60 = 2220

Part c) I'll list the first of each pattern, the number permutations of the pattern, the number of occurrences of such a pattern and the product:

(11,1,1,1,1) 5 * 2 = 10
(10,2,1,1,1) 20 * 9 = 180
(9,2,2,1,1) 30* 8 = 240
(8,3,2,1,1) 60 * 7 = 420
(6,6,1,1,1) 10 * 1 = 10
(5,4,3,2,1) 120 * 1 = 120
(3,3,3,3,3) 1 * 1 = 1
For a total of 981

Part d) Similar listing as used in c)
(9,3,2,1,0) 120*7 = 840
(12,2,1,0,0) 60*37 = 2220
(13,1,1,0,0) 30*13 = 390
(14,1,0,0,0) 20*20 = 400
(6,6,1,1,1) 10*2 = 20
(15,0,0,0,0) 5*3 = 15
(3,3,3,3,3) 1*1 = 1
For a total of 3886

Part e) Same as d) but fewer permutations count
(9,3,2,1,0) 60*7 = 420
(12,2,1,0,0) 27*37 = 999
(13,1,1,0,0) 12*13 = 156
(14,1,0,0,0) 7*20 = 140
(6,6,1,1,1) 3*2 = 6
(15,0,0,0,0) 1*3 = 3
(3,3,3,3,3) 0*1 = 0
For a total of 1724

 Posted by Jer on 2012-11-21 13:19:35

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