Q. Ten balls numbered from $1$ to $10$ are to be put in three boxes numbered $1,2 ,$ and $3$ so that no box remains empty, then the number in which this can be so that ' Numbers of balls of box $1$ > Numbers of balls of box $2$ > Numbers of balls of box $3$ ' is -
My Attempt- I made these cases for each box-
NO. OF BALLS-
$1) 7 , 2 , 1;$
$2) 6 , 3 , 1;$
$3) 5 , 4 , 1;$
$4) 5 , 3 , 2;$
Then, I calculated the total no. of cases using Permutation , But my Answer is coming way larger than the original Answer! I Dont know where I am doing some mistake! Pls help!
It looks like the problem has been interpreted to mean "number of balls in box 1..." whereas as phrased, it says "numbers on balls".
Looking at the problem as written, the answer is $\binom{9}{2}=36$.
Imagine lining up the balls in reverse order: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. You can construct each placement satisfying your criteria if you place 2 "dividers" in 2 of the 9 gaps between the balls, like 10, 9, 8, |, 7, 6, 5,|, 4, 3, 2, 1, and put the balls before the first divider in box 1, the balls between the dividers in box 2, and the balls after the 2nd divider in box 3. (Check that you've satisfied your criteria). There are $\binom{9}{2}$ ways to place these dividers.