Kenneth Rosen Edition 6th Exercise 6.6 Question 3 (Page No. 456)

Question

How many solution does the equation x1+x2+x3=13 have where x1  x2 and x3 are non negative less than 6 ?

Answer 1

Actually this is a question of multinomial theorem .The problem is similar to dice sum problem.The problem can be formulated mathematically as :

x₁  + x₂  + x₃ = 13 where each of the terms 0 <= x_i  <= 5  for i from 1 to 3..

So its solution is given by coefficient of x^r  in  (1 + x + x² + .................x^k) ⁿ .

Here k = 5 , r = 13 and n = 3   

So coefficient of x¹³  in  (1 + x  ... + x⁵)³   ,Now the inner term is :

1 + x  ..+ x⁵  which  is a G.P. which is =   (1 - x⁶) / (1-x)

So the given term (1 + x  ... + x⁵)³  can be rewritten as :   (1 - x⁶)³  .  (1 - x)^-3

                                                                              =   (1 - 3x⁶ + 3x¹² - x¹⁸).( 1 - x)^-3

Now since x¹⁸ is already larger than x¹³ , so we ignore it .Here we have to remember the result:

Coefficient of x^r in (1 - x)^-n   is given by ^n-1+r C_r.So

a) Taking 1 from the 1st term , we require r = 13 and n is 3 in this case.Therefore coefficient of x¹³ due to this term is given by :

  ^3-1+13C₁₃   =    ¹⁵C₁₃  =   15.14/2  =  105

b) Now taking -3x⁶ as the term , we need r = 7 and n = 3 .So the coefficient of x¹³ due to this term is given by :

(-3) .  ^3-1+7C₇   =   (-3) .  ⁹C₇  =   -3 . 36  =  -108

c) b) Now taking 3x¹² as the term , we need r = 1 and n = 3 .So the coefficient of x¹³ due to this term is given by :

(3) .  ^3-1+1C₁   =   (3) .  ³C₁  =   3 . 3   =   9

So combining a), b) and c) , we get number of required solutions as :  105 - 108 + 9   = 6

Comments

nice explanation @habib. (y)

In short,
x1+x2+x3=13 , where 0<= x1,x2,x3 < 6.

possible combination - 3 5 5    -> 3 ways

                                  4 4 5   -> 3 ways

                      total = 6 ways.

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Ya alternatively we can also do like this. I gave actually the generic method..:)