## Problem 12

**Problem – 12**

How many solutions are there to the equation?

x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=23,

where *x _{i}*,

*i*=1, 2, 3, 4, 5, is a nonnegative integer such that

**a)** *x*_{1}≥1?

**b)** *x _{i}*≥1 for

*i*=1, 2, 3, 4, 5?

**c) ****0≤ x_{1}≤7**?

**d) **0≤*x*_{1}≤3, 1≤*x*_{2}<4, and *x*_{3}≥15?

** **

**Solution :**

a) We require x_{1} ≥ 1. This uses up 1 of the 23 total required, so the problem is the same as fining the number of solutions to x_{1}’+x_{2}’+x_{3}’+x_{4}’+x_{5}’ = 22 with each x_{i}’ a nonnegative integer. The number of solutions are therefore

C(5+22-1,22) = C(26,22) = 14950

b) We require each x_{i}’ ≥ 1. Thus up 5 of the 23 total required, so the problem is the same as finding the number of solutions to x_{1}’+x_{2}’+x_{3}’+x_{4}’+x_{5}’ = 18 with each x_{i}’ a nonnegative integer. The number of solutions are therefore

C(5+18-1,18) = C(22,18) = 7315

c) The number of solutions without restriction is C(5+23-1,23) = C(27,23) = 17550. The number of solutions violating the restriction by having x_{1}≥8 is C(5+15-1,15) = C(19,15) = 3876. Therefore the answer is 17550 – 3876 = 13674

d) The number of solutions with x_{2} ≥ 1 and x_{3} ≥ 15 ( as required) but without the restriction x_{1}≤3 and x_{2}<4 is C(5+7-1,7) = C(11,7) = 330. The number of solutions violating the restriction for x_{1}, which is ≥ 0, by having x_{1}≥ 4 is C(5+3-1,3) = C(7,3) = 35, the number of solutions violating the restriction for x_{2} , which is ≥ 1, by having x_{2}≥4 is C(5+4-1,4) = C(8,4) = 70. Therefore the answer is 330 – 35 – 70 = 225