Number of ways to divide $160$ into three parts 
In how many ways can we divide $160$ into three parts, each with at least $40$ and at most $60$?

The amounts can only be integers. Please help me solve this, I don't really get exercises with at least and at most conditions.
 A: SOLUTION:
This can be very easily solved by Partition method and then using generating function (other name: INTEGRAL EQUATION METHOD).
Let $\displaystyle x+y+z = \$160 \ \ $ as each gets atleast $\$40$  so just divide $\$40$ to each of them before hand.
Then the equation becomes $$\displaystyle a+b+c = \$40 \ \ $$ for some $a,b,c$.
Now we need to divide $\$20$ among $a,b,c$ as at max each can get $\$60$. 
Thus By generating function or Integral equation method we can convert the following equation into this form :-
$$\displaystyle (1 + r^1 + r^2 + ... + r^{20})^3 = r^{40} $$ 
Where the coefficient of $ r^{40} $ determines the number of possible solutions to the equation. By GP-summation we convert this into:
$$\displaystyle \frac {(1-r^{21})^3}{(1-r)^3} = r^{40} \implies (1-r^{21})^3(1-r)^{-3} = r^{40} $$
Now if we ibserve it then we can see that the terms that matter to us in $({1-r^{21}})^3$ is just $({1-3r^{21}})$, as rest of the terms have exponents greater than $40$.
So, we get the coefficient of $r^{40}$ by $\displaystyle \binom {40 + 3-1}{3-1} - 3 \times \binom {19 + 3-1}{3-1}$, thus the answer is:
$$\displaystyle \binom {42}{2} - 3\binom {21}{2} = 231 .$$
A: So first, because each number is at least $40$, so we may minus 40 from each number. So the remaining part is how to divide $40$ into $3$ parts, each part is at most $20$.
We now consider $a,b,c$ with $a+b+c=40$, then set $x=20-a, y=20-b,z=20-c$, so $x+y+z$=$20$ and $x,y,z$ is at least $0$.
Now it is clear that the number of triples $(x,y,z)$ is the number of $(a,b,c).$
We may play the bar and star here. 
So let there be 20 stars on a row and we must place 2 bars to separate the three parts.
If all 3 numbers are positive then there are $\displaystyle \binom{19}{2} = 171 $ triples
If one of the three number is $0$ then there are $\displaystyle 3\times \binom{19}{1} =57 $ triples.
If one of the thee numbe is $20$ then there are $3$ triples.
In total there are $231$ triples.
A: The condition:
$$x'+y'+z'=160, \ 40\le x',y',z'\le 60$$
...is equivalent to:
$$x+y+z=40, \ 0\le x,y,z\le 20 \tag{1}$$
...with:
$$x'=x+40,\ y'=y+40,\ z'=z+40$$
So the number of soulutions that you are looking for is identical to the number of solutions of (1). Notice that we have decide about values of $x,y$. The value of $z$ can be calculated from (1).
$$x=0 \implies y=20\tag{2-0}$$
$$x=1 \implies y \in \{20, 19\}\tag{2-1}$$
$$x=2 \implies y \in \{20, 19, 18\}\tag{2-2}$$
$$x=3 \implies y \in \{20, 19, 18, 17\}\tag{2-3}$$
$$\dots$$
$$x=20 \implies y \in \{20, 19, 18, 17,\dots,0\}\tag{2-20}$$
So you have 1 solution from (2-0), 2 solutions from (2-1), 3 solutions from (2-2).... and 21 solution from (2-20). The total number of solutions is:
$$1+2+3+...+21=231$$ 
A: All of the parts are greater than or equal to $40$, 
So the sum will be at least $120$.
Now the remaining $40$ needs to be partitioned into $3$ groups $A,B,C$ so that none of the groups have more than $20$
The total ways to do this without the $20$ limit restriction is:
$A+B+C=40$, hence $\binom{42}{2}$ ways. 
Now we have to subtract ways in which  $1$ group has more than $20$ 
Lets say that $A \gt 20$
So $B+C \lt 40-20$, or in other words, $B+C+D*=19$
This can be done in:  $\binom{21}{2}$ ways
Now any of $A,B,C$ could have been $\gt 20$
$3\cdot \binom{21}{2}$
And finally, 
$\binom{42}{2}-3\cdot \binom{21}{2}$
*$D$ was added as another variable to account for the fact that if $B+C=18$, $D=1$, similarly if $B+C=12$, $D=7$ and so on
