What is the difference between these two combinations? Excerpt from "Probability Theory: A Concise Course", Y.A. Rozanov (Chapter 2).

A batch of 100 manufactured items contains 5 defective items. Fifty
  items are chosen at random and then inspected. Suppose the whole batch
  is accepted if no more than one of the 50 inspected items is
  defective. What is the probability of accepting the whole batch?

(A) My solution:
N: The number of combinations of 100 items taken 50 at a time is: ${100\choose50}$
N(A): The number of ways we can select 50 items out of 95 (non-defective) + 1 (one defective allowed) items or 96 items: ${96\choose50}$
The probability is then: $P(A) = \frac{96\choose50}{100\choose50}$
(B) Verified solution:
N: The number of combinations of 100 items taken 50 at a time is: ${100\choose50}$
N(A): The number of ways we can select 50 items out of 95 non-defective ones in addition to the number of ways we can select 49 items out of 95 non-defective ones and 1 out of 5 defective items: ${95\choose50} + {95\choose49}{5\choose1}$
The probability is then: $P(A) = \frac{{95\choose50} + {95\choose49}{5\choose1}}{100\choose50}$
My question:
I understand the logic of the verified solution B but I can't get how it is different from A. With my first solution am I not computing the number of ways of arranging 50 items out of 96 items (95 good and 1 defective)?
Can you please also give me in plain english what the solution A does?
Thank you all
 A: Because at no point in the process are you selecting 50 items from 96 options. You don't add to the top number in binomial coefficients like that.
Let $A$ be the event that none of the 50 are defected and let $B$ be the event exactly 1 is defective. We want the probability of the event $A \ \cup B $ (A or B). That is the probability that either none of the 50 are defective or exactly 1 is defective. Both are acceptable.
Because they are mutually exclusive, $P(A \ \cup B \,) = P(A) + P(B)$
The total number of ways of picking 50 items from 100 is $\binom{100}{50}$
There are 95 non-defective items.
$P(A)$:
Here you need to choose 50 items from the 95 non-defective ones.
There are $\binom{95}{50}$ ways to do this, so 
$$P(A) = \frac{\binom{95}{50}}{\binom{100}{50}}$$
$P(B)$
Here you use the rule of product. You can break down the task of choosing 50 items, 1 of which is defective, into choosing 49 non-defective items (out of 95) and choosing 1 defective item (out of 5).
There are $\binom{95}{49}$ ways of doing the first part and $\binom{5}{1}$ ways of doing the second.
Thus P(B) = $$\frac{\binom{95}{49}\binom{5}{1}}{\binom{100}{50}}$$
Altogether we have 
$$P(A \ \cup B) = P(A) + P(B) = \frac{\binom{95}{50}}{\binom{100}{50}} + \frac{\binom{95}{49}\binom{5}{1}}{\binom{100}{50}} = \frac{\binom{95}{50} + \binom{95}{49}\binom{5}{1}}{\binom{100}{50}}$$
