Split $151$ cakes amongst $3$ people $151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?
The solution is $${\binom {3+74-1}{3-1}}$$
which is exactly a combination with replacement. 
I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.
 A: The solutions of the system :
$$
x_1 + x_2 + x_3  = 151 : 0 \leq x_1,x_2,x_3 \leq 75 \\ 
$$
is in one-one correspondence with the solutions of the system :
$$
(75 - x_1) + (75 - x_2) + (75 - x_3) = 3 \times 75 - 151 : 0 \leq x_1,x_2,x_3 \leq 75 
$$
Which after the change of variable $y_i = 75-x_i$ becomes (since $\color{blue}{75\times3 - 151 = 74}$):
$$
y_1+y_2+y_3 = 74 : 0 \leq y_1,y_2,y_3 \leq 75
$$
which by non-negativity has the same solutions as :
$$
y_1+y_2+y_3 = 74 : 0 \leq y_i \forall i
$$
Now, play the stars-and-bars game to find the number of solutions of this equation.
A: "There are no coincidences in mathematics."
The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$\left(\!\!\!\binom{75}{2}\!\!\!\right)=\binom{75+2-1}{2}=\binom{74+3-1}{3-1}$$
This requires some explanation: There is only one way of combining the two numbers to give three numbers less than $75$. For example, if you choose $a$ and $b$, with $a\ge b$, and think of 75 as being split into $a+(75-a)$ and $b+(75-b)$; then combine them to give three acceptable numbers as $a+(75-a+b)+(75-b+1)=151$.
