Here, (x, y, z) means put x letters in the first box, y in the second, and z in the third.
(6, 0, 0)
(5, 1, 0)
(5, 0, 1)
(4, 2, 0)
(4, 1, 1)
(4, 0, 2)
(3, 3, 0)
etc.
Before long, you notice a shortcut and find an easier way to count, as there's 1 possible combination with 6 in the first box, 2 combinations with 5 in the first box, and so on. You conclude from that that if you had 100 letters, you could solve the problem just as easily.
Good work in advance.
2007-01-18 11:08:20
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answer #1
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answered by Anonymous
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distributions are:
6 0 0, 5 1 0, 5 0 1, 4 2 0, 4 0 2, 4 1 1, 3 3 0, 3 0 3
3 2 1, 3 1 2, 2 4 0, 2 0 4, 2 3 1, 2 1 3, 2 2 2, 1 5 0
1 0 5, 1 4 1, 1 1 4, 1 3 2, 1 2 3, 0 6 0, 0 0 6, 0 5 1
0 1 5, 0 4 2, 0 2 4, 0 3 3
or 28 possible ways
2007-01-18 19:22:34
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answer #2
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answered by Helmut 7
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1 way
2007-01-18 19:02:38
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answer #3
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answered by J.r. 3
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