YLL討論網

3. 已知X+Y+Z=10，且X≥1，Y≥2，Z≥3， 試求其整數解的組數。

    若只是要求非負整數解，只需要求( H 3取10 ) 的值就好。

    但題目要求 X≥1，Y≥2，Z≥3，
  
    所以 X先丟1個給它，Y先丟2個給它，Z先丟3個給它

    X' + Y'' + Z''' = 4  (共丟了6個，所以只剩4個)

    H 3取4 = C 6取4

               = C 6取2

               = (6 x 5) / (2 x 1)

               = 15

2. 已知將8的相同的球放進5個不同的箱子，若規定每個箱子至少放1個球，則有多少種放法?

   相同的東西，丟入不同的相子 => 重複組合

    X + Y + Z + W + U = 8

   而每個箱子至少放1個球，則是要求正整數解，所以先各丟1個給他們。

   X' + Y'+ Z' + W' + U' = 3 (各丟1個後，只剩下3個)

   H 5取3 = C 7取3

              = (7 x 6 x 5) / (3 x 2 x 1)

              = 35

1. 試求X+Y+Z+U=12的非負整數解及正整數解各有多少組?

   非負整數解：

   H 4取12 = C 15取12

                = C 15取3

                = (15 x 14 x 13) / (3 x 2 x 1)
    
                = 455

   正整數解：各丟給1之後再算

   X' + Y' + Z' + U' = 8  (各丟1個後，只剩8個)

   H 4取8  = C 11取8

               = C 11取3

               = (11 x 10 x 9) / (3 x 2 x 1)
    
               = 165

Since 2, 3 are similar, you should try and solve them yourself.

1.
picture this:
you are trying to find out the permutation of 12 rocks and 3 sticks (3 sticks divides the rocks into 4 spaces which are X,Y,Z,U respectively)
xxxxxxxxxxxx |||

eg.   xxxx|xxxx|xx|xx (X,Y,Z,U)

so the non negative sol would be  (12+3)!/(12!3!)

now if you want positive sol:
first assign each (X,Y,Z,U) 1 rocks each, then the problem will be reduced to similar to the one we did except that
the initial setup would then be the following:
xxxxxxxx, |||
so the you would be finding the permutation of 8 rocks  and 3 sticks.
find the permutation and you're done

1.試求X+Y+Z+U=12的非負整數解及正整數解各有多少組?



2.已知將8的相同的球放進5個不同的箱子，若規定每個箱子至少放1個球，則有多少種放法?



3.已知X+Y+Z=10，且X≥1，Y≥2，Z≥3， 試求其整數解的組數。