How many integer solutions?

First, apply the given conditions. We write $b=2j+1,c=2k, d=2m, e=a+3, g=f$ $\begin{aligned} a+b+c+d+e+f+g+h &= 10 \\ a+2j+1+2k+2m+a+3+2f+h&=10\\ 2(a+j+k+m+f+2)+h &=10 \implies h = 2n \\ 2(a+j+k+m+f+n+2) &= 10 \\ a+j+k+m+f+n+2 &= 5 \\ a+j+k+m+f+n &= 3\end{aligned}$ How many $6$ -tuples of non-negative integers sum to $3$ ? Exactly $\binom{6 + 3 - 1}{3} = \binom{8}{3}$ which comes out to $\boxed{56}$ .

Note: All numbers in the following solution are non-negative integers.

We know that $\begin{aligned}e&=&a+3,\\f&=&g.\end{aligned}$ Since $b$ is odd, it can be expressed as $2b'+1$ where $b'$ . Similarly, $c$ and $d$ can be expressed as $2c'$ and $2d'$ , respectively. Substituting all this into our original equation gives $\begin{aligned}a+b+c+d+e+f+g+h&=&10\\2a+2f+2b'+1+2c'+2d'+h+3&=&10\end{aligned}$ Taking both sides $\text{mod 2}$ gives $h\equiv0\pmod2$ so $h$ can be expressed as $2h'$ . We can substitute this and simplify further: $\begin{aligned}2a+2f+2b'+2c'+2d'+h&=&6\\a+f+b'+c'+d'+h'&=&3\end{aligned}$ The number of solutions to this equation is clearly $\binom83=\boxed{56}$ by balls and urns.

Let $k=\frac{b-1}{2},l=\frac{c}{2},m=\frac{d}{2}$ . Then the whole equation can be simplified into $2a+2k+2l+2m+2f+h=6$ which we get that $h$ is even. So it can be further simplified into $a+k+l+m+f+n=3$ , where $n=\frac{h}{2}$ . No of soln to $a+k+l+m+f+n=3$ is the no. of ways to put 2 sticks and 6 balls in a row. So $\binom{8}{3}$ is our desired answer.

We are given that there exist $b',c',d', \in \mathbb{N} \cup \{0\}$ $b = 2b' + 1$ , $c = 2c'$ , $d = 2d'$ , hence $2a + 2b' + 2c' + 2d' + 2f + h = 6$ This forces $2 \mid h$ , i.e. there exists $h' \in \mathbb{N} \cup \{0\}$ such that $h = 2h'$ , so $a + b' + c' + d' + f + h' = 3$ Let $a' = a + 1$ , $b'' = b' + 1$ , $c'' = c' + 1$ , $d'' = d' + 1$ , $f' = f + 1$ , $h'' = h' + 1$ , then $a' + b'' + c'' + d'' + f' + h'' = 9$ and $a', b'',c'', d'', f', h''$ are positive. The problem is readily seen to be equivalent with number of ways to separate $9$ stars with $5$ bars, as depicted in following figure $\star \mid \star \mid \star \star \star \mid \star \mid \star \star \mid \star$ note that there are $8$ gaps between the stars, hence the number sought is $\binom{8}{5} = \frac{6 \cdot 7 \cdot 8}{3!} = 7 \cdot 8 = \boxed{56}$

Since $e=a+3$ and $e=f$ , the given equation can be written as

$2a+b+c+d+2f+h=7$

Let $b=2k+1$ , $c=2m$ and $d=2n$ where $k,m,n \in \{ 0,1,2.... \}$ using the fact that $b$ is odd and $c$ and $d$ are even. Rewriting the equation,

$2(a+k+m+n+f)+h=6$

Let $a+k+m+n+f=x$ . The possible values of $x$ are $0,1,2$ and $3$ . For these values, the integer solutions can be easily calculated.

When $x=0$ , there is only 1 possible solution. For $x=1$ , number of solutions is 5. Similarly for $x=2$ and $x=3$ , the number of solutions are 15 and 35 respectively. Adding them up,

$1+5+15+35=\fbox{56}$

Since $b$ is odd

We can express $b$ as

$b=2k+1$ where $k$ is any non-negative integer.

Similarly, we can also express $c$ and $d$ as

$c=2m$ and

$d=2n$ where $m$ and $n$

are also non-negative integers.

Using these, together with the fact that

$e=a+3$

and

$f=g$ ,

we will get a new equation as

$a+2k+1+2m+2n+a+3+f+f+h=10$

which can be simplified to

$2a+2k+2m+2n+2f+h=6$ .

Now, $h$ can only have four possible values: $0, 2, 4$ or $6$

Case 1

If $h=0$ ,

The equation we got can be further simplified to:

$a+k+m+n+f=3$

The number of non-negative solutions for that equation is given by

${3+5-1 \choose 5-1}={7 \choose 4}=35$

Case 2

If $h=2$ ,

The equation we got can be further simplified to:

$a+k+m+n+f=2$

The number of non-negative solutions for that equation is given by

${2+5-1 \choose 5-1}={6 \choose 4}=15$

Case 3

If $h=4$ ,

The equation we got can be further simplified to:

$a+k+m+n+f=1$

The number of non-negative solutions for that equation is given by

${1+5-1 \choose 5-1}={5 \choose 4}=5$

Case 2

If $h=6$ ,

The equation we got can be further simplified to:

$a+k+m+n+f=0$

The number of non-negative solutions for that equation is given by

${0+5-1 \choose 5-1}={4 \choose 4}=1$

Taking the sum of all possible cases, we have

$35+15+5+1=56$ possible non-negative integer solutions.

Let:

$b = 2b'+1$ , $c=2c'$ , $d=2d'$ , where $b',c',d' \in [0,...]$ . Then,

$a+b+c+d+e+f+g+h=10$

$\implies 2(a+b'+c'+d'+f)+h=6$ .

Since $\{a,b',c',d',f\}$ are non negative, $h$ is even.

Case 1: $h=0$ . Then $a=0,b=1,c=0,d=0,e=3,f=g=0$ is a unique solution.

Case 2: $h=4$ . Then $a+b'+c'+d'+f=1$ . So any one of the 5 variables $a,b',c',d',f$ can be 1 and all the others are 0. This can be chosen in $5 \choose 2$ ways.

Case 3: $h=2$ . Then $a+b'+c'+d'+f=2$ . Then either one of the 5 variables is 2 and the rest are zero or any two of the 5 variables are 1 and the rest are zero. This can be chosen in $5 \choose 1$ + $5 \choose 2$ ways.

Case 3: $h=0$ . In this case, there are 3 possibilities.

a) One variable is 3 and the rest are zero. This can be chosen in $5 \choose 1$ ways.

b) One variable is 2, another variable is 1 and the rest are zero. This can be chosen in $2{5 \choose 2}$ ways.

c) Three variables are 1 and the rest are zero. This can be chosen in $5 \choose 3$ ways.

Add all these up, we have $1+{5 \choose 1} + {5 \choose 1} + {5 \choose 2} + {5 \choose 1} + 2{5 \choose 2}+{5 \choose 3}=56$ ways.