Question

How many solutions are there to the equation x1 + x2 + x3 + x4 + x5 = 21,

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

a) x1 ≥ 1?

b) xi ≥ 2 for i = 1, 2, 3, 4, 5?

c) 0 ≤ x1 ≤ 10?

d) 0 ≤ x1 ≤ 3, 1 ≤ x2 < 4, and x3 ≥ 15?

Short answer

Number of solutions are

1. $c(24,4)=10626$
   
2. $c(15,4)=1365$
   
3. $c(25,4)-c(14,4)=11649$
   
4. there are 106 non-negative integer solutions.

Step-by-step solution

Step 1: Use the formula for string factorial 

Total number of arrangements of n objects if al are different

If r! of them are same number of ways are given by

$C(n+r-1,r)=\frac{(n+r-1)!}{n!r!}$

The n! Is string length

r! are outcome of strings

Solution of x1 ≥ 1

Let’s, applied r! and n! values,

Here,

$\begin{array}{r}C(n+r-1,r)=\frac{(n+r-1)!}{n!r!}\\ n=21\\ r=4\\ C(n+r-1,r)=C(21+4-1,4)\\ c(24,4)=10626\end{array}$

Solution of xi ≥ 2 for i = 1,2,3,4,5

Let’s, applied r! and n! values,

Here,

$\begin{array}{r}C(n+r-1,r)=\frac{(n+r-1)!}{n!r!}\\ n=21\\ r=4\\ C(n+r-1,r)=C(21+4-1,4)\\ c(24,4)=10626\end{array}$

Solution of 0 ≤ x1 ≤ 10

Let’s, applied r! and n! values,

$\begin{array}{r}C(n+r-1,r)=\frac{(n+r-1)!}{n!r!}\\ n=21\\ r=4\\ C(n+r-1,r)=C(21+4-1,4)\\ c(24,4)=10626\\ n=11\\ r=4\\ C(n+r-1,r)=C(11+4-1,4)\\ c(14,4)=1001\\ c(25,4)-c(14,4)=12650-1001\\ =11649\end{array}$

Step 5: Solution of 0 ≤ x1 ≤ 3, 1 ≤ x2 < 4, and x3 ≥ 15

Here, let consider the condition

The coefficient of$x^{21}=106$ is your answer

hence there are 106 non-negative integer solutions.