Number of ways picking up shoes not belonging to a pair A closet contains 5 pairs of shoes.What is the number of ways in which 4 shoes can be chosen from it so that there will be no complete pair ?
My attempt : My attempt was to find out the total number of ways in which 4 shoes can be chosen out of ten and then subtract the number of ways in which at least one pair will be chosen that is $ 5* {8 \choose 2} $.
But this gives 70 as the answer , while the correct answer is 80. What am i doing wrong ?
 A: One way is to first use permutations:
The first shoe can be chosen in $10$ ways, but to avoid pairs, for the next there are only $8$ choices, then $6$ choices, and so on.
Thus permutations of $4$ "non-pairs" $= 10\cdot8\cdot6\cdot4$
But we are only interested in this collection, not their order, thus divide be $4!$
$\dfrac{10\cdot8\cdot6\cdot4}{4!} = 80$  
A: You can visualize it first: 

1 2 3 4 5
1 2 3 4 5

Now you can have a choice out of two in each column. For 4 pairs you choose first out of the pairs from number 1 to number 4. This $2^4=16$. We have 5 pair of shoes. The ways of choosing 4 out of 5 shoes of different pairs is ${5 \choose 4} =5$.  Therefore in total you have 80 ways of choosing 4 shoes of different pairs.
A: When subtracting the number of ways in which at least one pair is chosen, you double-counted the cases when you take exactly two pairs. You have to add again the number of ways in which you can take two pairs. Note that this number is ${5 \choose 2}=10$. So, you get your answer.
