Two problems on combinatorics Suppose we have a bag which has chips of four colors numbered $1$ to $13$, i.e. in total $52$ balls. Now what is the difference between these two problems.

Problem-$1$- In how many ways can you choose $3$ chips with one number and $2$ chips with a  different number?

My approach- First we have to calculate no. of ways to select two different numbers from $\{1,2,\dots, 13\}$ , which are $13\times 12$. Now suppose we chose number $2$ and $4$, then I have $^4C_3$ choices for chip number $2$ and $^4C_2$ choices for chip number $4$. So the answer is $12\times 13\times 4\times 6=3744$. The book also worked like this, thus same answer.

Problem-$2$- In how many ways can you choose $2$ chips with one number and $2$ chips with a different number, and one chip of an other number?

My doubt- Now should not be the approach here same like in first question, and thus we should get answer $13\times 12\times 11\times ^4C_2 \times ^4C_2 \times ^4C_1= 13\times 12\times 11\times 6 \times 6 \times 4 =247104$.
But the book says (on page $86$) that "in this problem there is no way to distinguish the difference between a pair of chips numbered C  and a pair of chips numbered D. In this problem you choose your chips at once. The chips comes to you at same time."
What is different in this problem, why do now suddenly chips come at same time. It is word to word same as previous problem just one more slot. The book calculates the answer as $^4C_2 \times ^4C_2 \times ^{13}C_2 \times 44=123552$ i.e. half of what I calculated.
Aren't both problems same?
 A: 
What is different in this problem, why do now suddenly chips come at same time.

If you count all the ways to choose two colours of a number, and then two colours of another number, you may be over counting, because it doesn't matter which order you choose pair of colours for two numbers.
Example: If you choose $\rm 1R, 1B$ then $\rm 2R, 2B$ is this any different from choosing $\rm 2R, 2B$ then $\rm 1R, 1B$?
However, choosing $\rm 1R, 1B, 1G$ then $\rm 2R, 2B$ is clearly different from choosing $\rm 2R, 2B, 2G$ then $\rm 1R, 1B$.   The distinction lies in the size of the selection for each number; so they can't be interchanged.
To see this in action, take four balls: $\rm\{1R, 1B, 2R, 2B\}$.   Now count the ways to select four of these four balls such that you have a pair of colours for two numbers.

 There can only be one.

A: The book answer is right, and it has already been explained why, 
but I would prefer to arrange the terms differently for greater conceptual clarity
$${13\choose 2}{4\choose2}^2 {11\choose 1}{4\choose 1}$$
