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3 answers

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 · answer #1 · answered by Anonymous · 0 0

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 · answer #2 · answered by Helmut 7 · 1 0

1 way

2007-01-18 19:02:38 · answer #3 · answered by J.r. 3 · 0 2

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