Question: > 
> The number of ways in which 5 different books can be distributed among 10 people if each person can get at most two books is?
> 
> 
> 

My Attempt:

The number of possibilities is  

$(2,2,1)$[rest all zero],$(1,1,1,1,1)$[rest all zero], $(2,1,1,1)$[rest all zero]

Therefore, dividing 5 books and distributing them will give me,

$\binom{10}{3}[\frac{5!}{2!2!1!2!}] + \binom{10}{5}[\frac{5!}{1!1!1!1!1!}] + \binom{10}{4}[\frac{5!}{2!1!1!1!}]$

This may be the answer according to me.

But I don't know the correct answer to the given question. So, can anyone check my work?

Answer: 
test case for {2,1,1,1}

no of ways to select 4 person($P\_1,P\_2,P\_3,P\_4$) =$\binom{10}{4}$

no of ways we can divide 5 books into 4 groups of {2,1,1,1}($G\_1,G\_2,G\_3,G\_4$)

=$\frac{5!}{2!1!1!1!}$

now we have to assign these ($G\_1,G\_2,G\_3,G\_4$) to ($P\_1,P\_2,P\_3,P\_4$) which can be done in 4 ways

Total = 4 $\binom{10}{4}$$\frac{5!}{2!1!1!1!}$ = 240 $\binom{10}{4}$ 

Method 2:

instead of dividing into groups lets calculate just how many ways we can choose 2 books out of 5 =$\binom{5}{2}$

Now these can be arranged to ($P\_1,P\_2,P\_3,P\_4$) in 4! ways 

Total = 4! $\binom{5}{2}$ $\binom{10}{4}$ = 240 $\binom{10}{4}$ 

You can choose any method you want , its just that in method one you didnt consider the matter of assigning

Also when @Tal-Botvinnik mention (2 over 5) and (2 over 3) he is choosing {2,2,1} by method 2
Your approach is right. But suppose we are looking at the case $(2,2,1)$, then once you have chosen three people over the ten, say people $p\_1,p\_2,p\_3$ you have to choose which one of the three receives 1 book (i.e. multiply by three), then you have to choose which books are to be given to person 1 (choose two over five) and to person 2 (choose two over three). This will cover all cases in this setting.

$${10\choose 3}\times 3\times {5\choose 2}\times{3 \choose 2}.$$

The other two cases $(2,1,1,1)$ and $(1,1,1,1,1)$ are similar, I'll leave you to it.