Principles of Mathematics in Operations Research

288 Solutions

Similarly,

p(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 +

x^6 p(x) = x^6 + x^5 +

1 — x^6

(1 - x^6 )p(x) = 1 + x + x^2 + x^3 + x^4 + x^5 =

1-x

Then,

h(x) = (1 - x^6 )^4 \p(x)}^4 = (1 - x^6 )^4 (l +x + x^2 +x^3 + x^4 + ---)^4 = k(x)l(x);

by the Binomial theorem

and by the multiset problem

3\ M\ /5\ 2 /6\ 3 /7\ 4.. /12

«*> - W

+

w*

+

ur

+

w

1 +

Ur

• • •

+

1 • r

The ninth convolution of k(x)l(x) is the answer:

'J) C.

a

)-(0(0-»-"»'—

Therefore, the probability is

140

P(having a sum of 13) = -TJ- = 0.1080247

d)

an - 5an_i + 6an_ 2 = 0, Vn — 2,3,4,... <=$

anxn - 5a„_ixn + 6an_ 2 zn = 0, Vn = 2,3,4,...

Summing the above equation for all n, we get
oo oo oo
^2 anxn - 5 ^2 an-ixn + 6 ^ a„_ 2 a;" = 0
n=2 n=2 n—2

\g{x) - axx - a 0 ] - 5x[g(x) - a 0 ] + 6x^2 [g(x)} = 0

Using the boundary conditions (do = 2 and ai — 5) we have

,. ao + a\x — 5a,QX 2 — 5x 1 1

6x^2 -5x + l (3x-l)(2x-l) l-2x 1 - 3x

g(x) = (1 + 2x + 4x^2 + • • • + 2V + •••) + (1 + 3x + 9x^2 + ••• + 3V + •••)