Count the ways to select 12 coins from 5 identical bins There are five identical bins contains 4, 9, 7, 10, and 3 coins. How many ways to select 12 coins, such that at least a coin selected from each bins?
I don't have an idea. But, i thinks it is related to integer partition of 12 (with 5 parts).
At a glance, this question looks similar to other question asked by another user. But if you look at it carefully, you will notice that there is a big difference. It is the bins, here, it is identical.
 A: This is quite an unusual and question that seems somewhat pointless; I suspect that there’s a linguistic mistake somewhere along the way.
As noted in a comment, since you need to take at least one coin from each bin, you’re effectively taking $7$ coins from bins with $3$, $8$, $6$, $9$ and $2$ coins. Since you unconventionally don’t care about which bin which amount of coins was taken from, any partition of $7$ with at most $5$ parts can be realized by ordering both the parts and the bins and taking the largest part from the most copious bin, etc.; this works since the $k$-th bin has at least as many coins as the $k$-th part can have. There are $15$ partitions of $7$ and only two of them have more than $5$ parts, so there are $13$ ways to take the coins from the bins in the way that you want to count them.
On a side note, you count the ways, not the number of ways.
A: Almost every question posted in this field gives rise to a chain of comments
requesting / providing clarifications, but which normally do not get the goal.   
That is due to waiving to specify all the basic assumptions required to give an univoque answer, which are
 - in "common language", how the picking process is actually performed, and how the outcomes are
subsequently classified: considered / excluded, equivalent/different;
 - in classic probability terms, which is the space of equi-probable events considered;
 - in axiomatic terms, sample space & $\sigma$-algebra & probability measure.
Let's take a simplified version of your problem so that we can see what needs to be clarified 
to obtain a pertinent answer.
Let's have two bins, containing $2$ and $4$ undistinct coins. We go and pick, casually, three coins.
"Usually", the above process of picking-up the coins is understood as that we choose one of the two
bins, equi-probably, pick a coin from that and repeat independently from previous choice(s).
But it is clear that many other different picking processes can be devised
That leads to the binary tree sketched below.

It is clear that the equi-probable, and possible, events of picking $3$ coins from the two bins 
are the $7$ events in the last column. Out of these, $6$ (the whites) will translate into picking from both bins.
It is true that the $6$ events can be collected under the unique partition of $3$ into two parts, i.e.  $[2,1]$,
and which corresponds, apart permutation, to the  coins missing from the baskets.
But this "composed" event will not then be equi-probable with those discarded, and you are going to
miss or mis-compute the term "out of N total ways".
Coming finally to your real problem, and supposing that the correct understanding of the process  be as above,
since three of the bins will have a content below $7$,  it is difficult to provide a closed analytical formula, simpler than the following.
The picking process corresponds to the multinomial expansion
$$
\eqalign{
  & \left( {x_{\,1}  + \,x_{\,2}  + \, \cdots \, + x_{\,m} } \right)^{\,n}  =   \cr 
  &  =  \cdots  + \left. {\left( {x_{\,j_{\,1} }  \cdot \,x_{\,j_{\,2} }  \cdot \,\, \cdots \,\; \cdot x_{\,j_{\,n} } } \right)\;}
  \right|_{\;j_{\,k}  \in \left[ {1,m} \right]}  +  \cdots  =   \cr 
  & \sum\limits_{\left\{ {\matrix{
   {0\, \le \,k_{\,j} } \hfill  \cr 
   {k_{\,1}  + \,k_{\,2}  + \, \cdots \, + k_{\,m}  = \,n} \hfill  \cr 
  } } \right.} {\left( \matrix{  n \cr   k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,m}  \cr}  \right)
  x_{\,1} ^{\,k_{\,1} } \,x_{\,2} ^{\,k_{\,2} } \, \cdots \,x_{\,m} ^{\,k_{\,m} } }  \cr} 
$$
Putting the data and the capacity limits of your problem gives:
$$ \bbox[lightyellow] {  
N = \sum\limits_{\left\{ {\matrix{
   {0\, \le \,k_{\,j} }  \cr 
   {k_{\,1}  \le \,2,\;\,k_{\,2}  \le \,3,\;\,k_{\,2}  \le \,6}  \cr 
   {k_{\,1}  + \,k_{\,2}  + \, \cdots \, + k_{\,5}  = \,7}  \cr  } } \right.}
 {\left( \matrix{  7 \cr   k_{\,1} ,\,k_{\,2} ,\, \cdots \,,k_{\,5}  \cr}  \right)} 
 }$$
