In how many ways can you distribute 60 balls among 3 people such that the number of balls that C has is more than B and B has more than A.
(i) Can this question be rephrased as in a group of 60 people, A,B,C are a few. In how many of the total arrangements does C come before B and B comes before A ( left to right or right to left doesn't matter right?) ?
So my approach for the original question is as follows:
In 1/3rd of total arrangements C will have more balls than A and B , and in 1/2 od total arrangements B will have more balls than A. So the answer is 1/6th of total arrangements.
Total arrangements is non-negative solutions of a+b+c=60, which comes out to be choosing 2 from 62 I.E. $62 \choose2 $
So the answer is $\frac{62 \choose 2}{6}$ , but the answer so obtained is wrong. The actual answer is 300.
Can you guide me along the right process of thinking?
Let us define three new variables as below:
The number of balls A has = $x$
The number of balls B has = $x+y+1$
The number of balls C has = $x+y+z+2$
We are going to find the number of solutions to $3x+2y+z+3=60$ in which $x,y,z$ are bigger or equal to zero. Actually, we have $3x+2y+z=57$.
First, suppose $z=3m$ so $y$ must be $3n$, so we will have $x+2n+m=19$. We can put numbers instead of $n$. The number of answers in this case follows $20+18+...+2=110$.
Second, suppose $z=3m+1$ so $y$ must be $3n+1$, so we will have $x+2n+m=18$. We can put numbers instead of $n$. The number of answers in this case follows $19+17+...+1=100$.
Third, suppose $z=3m+2$ so $y$ must be $3n+2$, so we will have $x+2n+m=17$. We can put numbers instead of $n$. The number of answers in this case follows $18+16+...+2=90$.
Therefore, the total number is $110+100+90=300$.