text: # How many ways can we distribute $7$ different color pencils to $10$ drawers so there are at least $2$ pencils in the $10$th drawer? [closed]
  
  Тhere are $10$ drawers and $7$ different color pencils. How many ways we can arrange them, so there are at least $2$ pencils in the $10$th drawer?
  
  So i tried doing it this way:
  
  1.If all 7 pencils are in the 10 drawer this is 1.
  
  2.If 6 pencils are in the 10 drawer, this are Variations with rep. $\to 9^1=9$
  
  3.If 5 pencils are in the 10 drawer, this are Variations with rep. $\to 9^2=81$
  
  4.If 4 pencils are in the 10 drawer, this are Variations with rep. $\to 9^3=729$
  
  5.If 3 pencils are in the 10 drawer, this are Variations with rep. $\to 9^4=6561$
  
  6.If 2 pencils are in the 10 drawer, this are Variations with rep. $\to 9^5=59049$
  
  So when i sum them i get $66430$ but it's far from the answer.
  
  Any hints? Thank you!
  
  ## closed as off-topic by Austin Mohr, Magdiragdag, Claude Leibovici, Juniven, user223391 Apr 22 '17 at 13:52
  
  This question appears to be off-topic. The users who voted to close gave this specific reason:
  
  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Austin Mohr, Magdiragdag, Claude Leibovici, Juniven, Community
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  • Hint: Substract the conditions with 0 pencil or 1 pencil on the 10th drawer from the all combinations. – offret Apr 15 '17 at 17:16
  • A different approach from Burak's: how many ways can you distribute 5 pencils to 10 drawers, without additional restrictions? And how many ways can you pick 2 pencils out of 7? – Paul Sinclair Apr 15 '17 at 22:52
  • @PaulSinclair: That can work, but OP would have to be careful of overcounting. For instance, if pencils 1, 2, and 3 are in the tenth drawer, then without being careful, that would be counted as selecting pencils 1 and 2 to go in the tenth drawer, plus selecting pencils 1 and 3 to go in the tenth drawer, plus selecting pencils 2 and 3 to go in the tenth drawer. I think it would be harder to keep track of the overcounting than some more direct method. – Brian Tung Apr 21 '17 at 14:48
  
  There are $10^7$ ways of distributing seven pencils into ten drawers. Of course, many of those ways don't have at least two pencils in the tenth drawer. So let's subtract those out.
  
  How many ways are there to have no pencils in the tenth drawer? That's equal to the number of ways to distribute seven pencils into the first nine drawers, which is $9^7$. So we subtract that out.
  
  Now, how many ways are there to have exactly one pencil in the tenth drawer? Well, let's take it slowly: How many ways are there to have only the first pencil in the tenth drawer? That means that the remaining six pencils have to all be distributed into the first nine drawers, which is $9^6$. So we have to subtract that out. But that's just the first pencil. Each of the other six pencils can be the unique pencil in the tenth drawer, too. So that's six more subtractions of $9^6$.
  
  Altogether, then, we have
  
  $$10^7-9^7-(7 \times 9^6)$$
  
  which is, indeed, quite a ways from $66430$.
  
  The reason why you didn't get the right answer is that you didn't account for all the different combinations of pencils that could be in the tenth drawer. For instance, you write, for $4$ pencils in the tenth drawer, a total count of $729$. But that only takes care of the ways to put the other $3$ pencils in the other $9$ drawers. You forgot that there are many ways to select the $4$ pencils that will go into the tenth drawer—$\binom{7}{4} = 35$, in fact.
  
  If you take that into account in your approach, you will have
  
  $$1+\binom76\times9+\binom75\times9^2+\binom74\times9^3+\binom73\times9^4+\binom72\times9^5$$
  
  and you will end up with the same answer as above.
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