Saturday Puzzle 2
May 20, 2006 pm31 6:03 pm
Today is a three-puzzle Saturday. Answers will be up, posted by readers or by me, by next Friday. Here's #2:
I have 3 red handkerchiefs and 6 white ones, and plan on sewing them into one super 3 by 3 hankie. How many ways can I do this?
Let's think:
| W | W | W |
| R | W | W |
| W | R | R |
and
| W | W | W |
| W | W | R |
| R | R | W |
and
| W | W | R |
| W | W | R |
| W | R | W |
and
| W | R | R |
| R | W | W |
| W | W | W |
are really all the same hankie, just flipped over or turned around.
2 Comments
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JD2718 
Nice site you’ve got here! (I came via JennyD.) I count 16 distinct ways: if you number your square
123
456
789
they are 123, 456, 126, 125, 124, 129, 128, 127, 245, 345, 135, 137, 138, 159, 168, and 246. The check is that we have a total of 9C3 = (9*8*7)/(3*2*1) = 84 choices for three red hankies among 9, and you just have to count the symmetries of each arrangement as you go and make sure the total works out to 84. (I suppose this doesn’t guard against putting two versions of some arrangement twice and leaving out another, but it does check for missing exclusive-or extra possibilities.)
Nicely done. And thanks for the kind words.
My solution is non-combinatorial. I considered cases by corners C, edges E, and middle M. (map to your notation in paren)
EEE – 1 case (246)
EEC – 4 cases (124, 126, 128, 168)
ECC – 4 cases (123, 127, 129, 138)
CCC – 1 case (137)
CCM – 2 cases (135, 159)
CEM – 2 cases (125, 156