I am trying to solve some exercises. Those exercises have to do with P(n,r) and C(n,r).There are
exercise 1: How many ways we can load $100$ indistinguishable boxes into $3$ cars A,B,C,if every car can be loaded with $20$ to $40$ boxes?
How I solve it: It says same boxes, so step 1 I know that it is $C(n,r)$. We know that $n=100$, we know that $r=3$.
We can say that we can split into categories for A,suppose we use $40$ to load so n=100 and $r=100-40=60$, so $r=60$. $C(100,60)$
With the same way I can do the B, and I have for B the $C(60,20)$.
As a result the last C, $60-40$ car can be $C(20,0)$.
formula for A : I have $$C(n,r)= \frac{n!}{r!(n−r)!} = \frac{60!}{100!(100-60)!}= \frac{60!}{100!40!} =$$ Something doesn't go well.
formula for B : $\dfrac{60!}{60!(60!-40!)}$, which is wrong and this one.
formula for C : $\dfrac{20!}{20!(20!-0!)}=1$
Boxes $A,B,C$ means labeled boxes, so the problem boils down to putting indistinguishable balls into distinguishable boxes, and the tool to use is stars and bars
We simplify the problem in two steps so that we can apply the basic stars and bars formula
We first place $20$ balls in each box so the problem now reduces to placing $40$ balls in the three boxes with a (now) upper limit of $20$ per box
The integer equation is $x_1 + x_2 +x_3 = 40,\; 0\leq{x_i}\leq 20$
The second simplification is a change of variable, We write $(20-y_1) + (20-y_2) + (20-y_3) = 40,$ which gives $y_1+y_2+y_3 = 20, y_i \geq 0$
Now we can use the basic stars and bars formula, to get $\binom{20+3-1}{3-1} $