Distributing $8$ different articles among $7$ boys 
Problem Statement:-
Find the number of ways in which $8$ different articles can be distributed among $7$ boys, if each boy is to receive at least one article.


Attempt at a solution:-
First, start by numbering the boys from $1$ to $7$, which can be done in $7!$ ways.
After the boys have been numbered then we resume our task of distributing the articles to the boys. As, each boy is supposed to at least get a single article.
Hence,
$\text{Ways of giving $1$ article to the $1^{\text{st}}$ boy= $8\choose1$}\\
\text{Ways of giving $1$ article to the $2^{\text{nd}}$ boy= $7\choose1$}\\
\text{Ways of giving $1$ article to the $3^{\text{rd}}$ boy= $6\choose1$}\\
\vdots\\
\text{Ways of giving $1$ article to the $7^{\text{th}}$ (and the last) boy= $2\choose1$}$
Now remains the last article, which can go to any one boy of the $7$ boys in ${{7}\choose{1}}=7$ ways
So, the number of ways of distributing $8$ articles among $7$ boys$=$
$$7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}$$
But, the arrangement of the two things that the $1^{\text{st}}$ boy gets is of no relevance so we gotta remove the consideration of the two objects being permuted among themselves. 
Hence, the total the number of ways of distributing $8$ articles among $7$ boys$=$
$$\frac{7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}}{2!}$$
The textbook gives the answer to the question as 
$$7\times \left( \binom{8}{2} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} \times \binom{1}{1}\right)$$
Where am I going wrong

The main problem here is that you have ordered the boys and also regarded the distribution of the presents as extra options. 
Probably the simplest way to get to the answer is to select one boy to get two presents ($7$ options), he chooses those two ($\binom 82$) then each boy chooses in turn ($6!$ options) :
$$7\binom 82 \binom 61 \binom 51 \binom 41 \binom 31 \binom 21 \binom 11 = 7\binom 82 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 7!\binom 82 $$
Another way to view the last formulation is that you pick two presents to "bundle", then distribute the 7 packages to the 7 boys.
