MATH 351 Fall 2009 Homework 1 Due: Wednesday, September 30

Again we proceed in the same fashion and suppose that the Host is at 12 o’clock and
that the hostess is seated at 6 o’clock. Thus we must just order the remaining 10
people and there are 10! = 3628800 ways to do so.
Problem 11 How many integer solutions of:
x 1 + x 2 + x 3 + x 4 = 30
satisfy, x 1 ≥ 2, x 2 ≥ 0, x 3 ≥ −5, and x 4 ≥ 8?
To solve this problem we re-scale the variables by letting y 1 = x 1 − 2, y 2 = x 2 ,
y 3 = x 3 + 5, and y 4 = x 4 − 8. Solving for each x i and substituting into the equation
above we obtain:
(y 1 + 2) + y 2 + (y 3 − 5) + (y 4 + 8) = 30,
where each y i is now greater then or equal to 0. Manipulating the constants gives us:
y 1 + y 2 + y 3 + y 4 = 25
. As per definition in class, the number of integer solutions to this equation is given
by: ( ) ( )
25 + 4 − 1 28 28 · 27 · 26
= = = 3276.
4 − 1 3 3!
8