Summation over 7 variables? Piece of Cake.

The question is asking for the number of non-decreasing sequences of 7 numbers from 0 to 8.
Define $A = a + 1, B = b + 2, C = c + 3, \ldots, G = g + 7$ .
This transforms a non-decreasing sequence of 7 numbers from 0 to 8, to a strictly increasing sequence of 7 numbers from 1 to 15.

How many ways are there to pick a strictly increasing sequence of 7 number out of 15? Well, if we were to choose any 7 numbers, then there is a unique arrangement into an increasing sequence. Hence, the answer is ${ 15 \choose 7 } \times 1 = 6435$ .

The given connected summation can be divided into following summations :

Along with inequality , consider permutations of:

0 equality b/w $a$ and $g$

1 equality b/w $a$ and $g$

2 equalities b/w $a$ and $g$

3 equalities b/w $a$ and $g$

4 equalities b/w $a$ and $g$

5 equalities b/w $a$ and $g$

6 equalities b/w $a$ and $g$

Adding the above cases we get,

${9 \choose 7}+(6\times{9 \choose 6})+(15\times{9 \choose 5})+(20\times{9 \choose 4})+(15\times{9 \choose 3})+(6\times{9 \choose 2})+{9 \choose 1} = 6435$

$\Rightarrow Ans: 6435$ $\ddot\smile$

P.S. I am not good at explaining.

First I am going to prove the Hockey-Stick identity.
Lemma:
$\dbinom{r}{r} + \dbinom{r+1}{r} + \dbinom{r+2}{r} + \ldots \dbinom{n}{r} = \dbinom{n+1}{r+1}$

Proof :
$\dbinom{r}{r} = \dbinom{r+1}{r+1}$
$L.HS = \dbinom{r}{r} + \dbinom{r+1}{r} + \dbinom{r+2}{r} + \ldots \dbinom{n}{r} = \dbinom{r+1}{r+1} + \dbinom{r+1}{r} + \dbinom{r+2}{r} + \ldots \dbinom{n}{r}$
Using the property,
$\dbinom{n}{m} + \dbinom{n}{m-1} = \dbinom{n+1}{m}$
$L.H.S = \dbinom{r+2}{r+1} + \dbinom{r+2}{r} + \ldots \dbinom{n}{r} = \dbinom{r+3}{r+1} + \ldots \dbinom{n}{r}$
Using this repeatedly we get,
$L.H.S = \dbinom{n+1}{r+1}$
Q.E.D
Moving on to the question,
The sum can be written as,
$S =\displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\sum_{e=0}^{f}\sum_{d=0}^{e}\sum_{c=0}^{d}\sum_{b=0}^{c}\sum_{a=0}^{b}1$
$S = \displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\sum_{e=0}^{f}\sum_{d=0}^{e}\sum_{c=0}^{d}\sum_{b=0}^{c}(b+1)$
$\displaystyle \sum_{b=0}^{c}(b+1)$ the sum of first $(c+1)$ integers and is given by $\dfrac{(c+1)(c+2)}{2} = \dbinom{c+2}{2}$
$\therefore S = \displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\sum_{e=0}^{f}\sum_{d=0}^{e}\sum_{c=0}^{d}\dbinom{c+2}{2}$
Now using our lemma,
$\therefore S = \displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\sum_{e=0}^{f}\sum_{d=0}^{e}\dbinom{d+3}{3}$
$S = \displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\sum_{e=0}^{f}\dbinom{e+4}{4}$
$S =\displaystyle \sum_{g=0}^{8}\sum_{f=0}^{g}\dbinom{f+5}{5}$
$S = \displaystyle \sum_{g=0}^{8}\dbinom{g+6}{6}$
$S = \dbinom{8+7}{7} = \dbinom{15}{7}$
$\therefore S = \dfrac{15!}{7!8!} = 6435$
I know there's a better combinatoric approach, I just preferred the algebraic one!