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 20121121 17:24:02 