# Integer solutions of $x_1+x_2+x_3+x_4+x_5+x_6=19$ if every $x_i \ge 2$

What is the number of solutions of $$x_1+x_2+x_3+x_4+x_5+x_6=19$$ if every $x_i \ge 2$ and all $x_i$ are integers?

I know how to compute the number of solutions when every $x_i \geq 0$, but how do I do this for $x_i\geq 2$?

Hint: Let $x_i = y_i+2 \to y_1+y_2+y_3+y_4+y_5+y_6 = 7, y_i \in \mathbb{Z_{\geq0}}$. The number of solutions to this and then the original equation is quite well-known. In fact, it is: $\binom{7+6-1}{6-1}= \binom{12}{5}$.

• How it is done I have no idea : ( Jan 14, 2015 at 13:46
• See here for the partition function and then use combinatorics to find the number of ways to order each partition. Jan 14, 2015 at 13:56

For each $x_i$ on the left hand side, put $x_i= y_1 + 2,\;\;y_i \geq 0.\,$. Then ($y_i = x_i-2$.

That gives you the equation $$(x_1 -2) + (x_2-2) + \cdots +(x_6 - 2) = 19 - (6\cdot 2)$$ $$= y_1 + y_2+y_3+y_4+y_5+y_6 = 19 - (2\cdot 6) = 7$$

Now use the stars-and-bars (Theorem 2) for the number of non-negative solutions:

For any pair of natural numbers $n$ and $k,$ the number of distinct $k$-tuples of non-negative integers whose sum is $n$ is given by the binomial coefficient $$\binom{n + k - 1}{n}.$$

Here, we have $k = 6$ and $n = 7$, which means the number of non-negative solutions to the equation is given by: $$\binom{7+6-1}{7} = \binom{12}{7} = \binom{12}{5} = \dfrac{12!}{7!5!} = 792$$

• Note that $$\binom{12}{7} = \dfrac{12!}{7!5!} = \binom{12}{5}$$ Jan 14, 2015 at 14:12

Elaborating on the idea that $x_i=y_i+2$

You are looking for the number of integer partitions with six summands. We know there are 15 ways of partitioning the number 7 by the partition function Listing them out as $6$-tuples yields

$$(7,0,0,0,0,0), (6,1,0,0,0,0)$$ $$(5,1,1,0,0,0),(5,2,0,0,0,0)$$ $$(4,1,1,1,0,0),(4,2,1,0,0,0)$$ $$(4,3,0,0,0,0),(3,1,1,1,1,0)$$ $$(3,2,1,1,0,0),(3,2,2,0,0,0)$$ $$(3,3,1,0,0,0),(2,1,1,1,1,1)$$ $$(2,2,1,1,1,0),(2,2,2,1,0,0)$$

(Note there are only 14 here but $\pi(7)=15$. This is because the last partition is $(1,1,1,1,1,1,1),$ which is a $7$-tuple, not a $6$-tuple.)

Now each of these $6$-tuples can be counted accordingly. For example then number of ways to write $(3,2,2,0,0,0)$ is $\frac{6!}{2!3!}$ since 2 occurs twice and $0$ occurs 3 times in the tuple. Applying this to all 14 partitions yields

$$6!\left(\frac{2}{5!}+\frac{4}{4!}+\frac{1}{3!}+\frac{6}{2!3!}+\frac{1}{2!2!}\right)=792$$