Choosing 3 out of 5 balls--> 5c3
Choosing 1 out of 2 balls --> 2c1
Choosing 1 out of 1 --> 1c1
Now for a particular combination (b1 b2 b3 , b4, b5)
Main point: For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways ?
So, 5c3 * 2c1 * 1c1 * 3! . What is so hard to understand?
What is the mistake that am i not able to catch?
The mistake, again, is in including both the 2c1 and the 3!.
You choose 3 balls to go in the triple box (10 ways).
You choose 1 ball to go in the FIRST single box (2 ways).
You choose 1 ball to go in the SECOND single box (1 way).
At this point, you have to simply pick WHICH box will hold the 3 balls, which can be done in 3 ways. You DON'T also have to pick which of the single balls will go in which other box, because their order has already been picked. So the total number of ways is 10*3*3 = 60.
You could, instead, have just chosen the 3 balls (10 ways), and left it so you had a group of 3 and two (unordered) single balls, and then arranged those three items (3! = 6 ways). This gives 10*6 = 60 ways, again.
But you can't order the single balls, and then order them AGAIN as part of the whole set; that results in doubling the count, and is therefore wrong.
Now, let's look at what you did through one example. Suppose the balls are ABCDE. We choose three of them, say ACE to share a box, then one to be alone, say D, and another to be alone, B. One of the arrangements we can make of these is D, ACD, B; another is B, ACD, D. But these two arrangements would also have resulted from choosing ACD, then B, then D. That is where you have double counted.
Combinatorics is inherently difficult because of the ease with which we can miss details like overcounting. I always feel more confident when I can get the same result two different ways. You should never assume that an answer you find is right, without carefully thinking about whether you caught everything. (And that implies you should not insist you are right when someone else with more experience tells you you are wrong, and shows a better way.) The method both JeffM and I have explained, taking a series of steps to make a choice, is a good one, but even so it needs to be checked.
By the way, here is how pka's formula is derived, using your specific problem as an example:
We can put 5 distinct items in 3 distinct boxes in 3^5 = 243 ways.
We need to subtract the ways in which one box would be empty; there are 3 ways to choose the box in 3 ways, and then put the 5 items in the remaining 2 boxes in 2^5 ways, for a total of 3*32 = 96 ways.
But in subtracting that, we are double-counting ways in which two boxes would be empty, so we have to add that back on. There are 3 ways to choose which two boxes are empty, and 1 way to put all 5 items in the one remaining box, for a total of 3*1 = 3 ways.
So our result is 243 - 96 + 3 = 150.
The formula [MATH]\sum\limits_{k = 0}^N {{{( - 1)}^k}\dbinom{N}{k}{{(N - k)}^M}}[/MATH] gives
[MATH]\sum\limits_{k = 0}^3 {{{( - 1)}^k}\dbinom{3}{k}{{(3 - k)}^5}} =\\ {{{( - 1)}^0}\dbinom{3}{0}{{(3 - 0)}^5}} + {{{( - 1)}^1}\dbinom{3}{1}{{(3 - 1)}^5}} + {{{( - 1)}^2}\dbinom{3}{2}{{(3 - 2)}^5}} + {{{( - 1)}^3}\dbinom{3}{3}{{(3 - 3)}^5}} =\\ 1\cdot 3^5 - 3\cdot 2^5 + 3\cdot 1^5 - 1\cdot 0^5 = 243-96+3-0 = 150[/MATH]
which is what I just did.
Now, the work of thinking through it, and even perhaps the work of calculating, is a little harder this way, but if you did this for a living (and were sure when the formula applies!), you would want to use the formula. If what you want is to learn how to think clearly, what we've shown you is better.
