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

First (d):

With 15 bars and 5 partitions between what's given to each of the six children, the answer to (d) is C(20,5)=15504.

Then (c):

After giving one bar to each child, it's a question of dividing 9 bars among the six kids (9 bars and 5 partitions): C(14,5)=2002.

For the rest:

FOR a = 0 TO 15
FOR b = 0 TO 15 - a
FOR c = 0 TO 15 - a - b
FOR d = 0 TO 15 - a - b - c
FOR e = 0 TO 15 - a - b - c - d
 f = 15 - a - b - c - d - e
 IF a <> b AND a <> c AND a <> d AND a <> e AND a <> f THEN
 IF b <> c AND b <> d AND b <> e AND b <> f THEN
 IF c <> d AND c <> e AND c <> f THEN
 IF d <> e AND d <> f THEN
 IF e <> f THEN
   parta = parta + 1
 END IF
 END IF
 END IF
 END IF
 END IF

 totb = (a <> b) + (a <> c) + (a <> d) + (a <> e) + (a <> f)
 totb = totb + (b <> c) + (b <> d) + (b <> e) + (b <> f)
 totb = totb + (c <> d) + (c <> e) + (c <> f)
 totb = totb + (d <> e) + (d <> f) + (e <> f)
 totb = ABS(totb)
 IF totb = 14 THEN partb = partb + 1

 IF a > b THEN parte = parte + 1
NEXT
NEXT
NEXT
NEXT
NEXT
PRINT "a"; parta
PRINT "b"; partb
PRINT "e"; parte

finds

part count
a    720
b    4680
e    6672

Posted by Charlie on 2012-11-21 17:24:02