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

You have permuted both the boys and the articles, which is incorrect.
I'd prefer to work it out as
[ Choose "lucky" boy ] $\times$ [ Permute articles ] 
$= \dbinom71 \times \dfrac{8!}{2!1!1!1!1!1!1!}$
which, of course, could be abbreviated as $  7\times \dfrac{8!}{2!}$
PS:
On closer scrutiny, I find that the book answer you give is also incorrect !
