How many $6$-digit numbers contain exactly two 1's, two 2's and two 3's I assume that this requires a combinatorial, so I think that what I need to do $2\choose 6$ since if we pick the first few the first two digits to be $1$'s and the second two digits to be $2$ we only need to account for the last set of two.
In this case the answer would be $15$
The book I'm using says the answer is $90$. I don't understand this. 
In the following example provided in the book 

"A student is trying to remember a classmate’s $7$-digit telephone number. He recalls that the ﬁrst digit is $7$ and then there are three $1$'s and three $3$'s in some order. How many telephone calls would he have to make to be certain that he ﬁnally calls his friend’s telephone number?"

the answer is $20$ and is found by doing $3\choose 6$ since the first digit is know and once you've place the $1$'s you need only worry about the placement of the $3$'s. 
How are the two questions different? How do I solve the first one?
 A: If a six digit number contains exactly two $1$'s, two $2$'s, and two $3$'s, we can choose the positions of the $1$'s in $\binom{6}{2}$ ways.   We can select the positions of the two $2$'s from the four remaining positions in $\binom{4}{2}$ ways.  We can select the position of the two $3$'s from the two remaining positions in $\binom{2}{2}$ ways.  Thus, the number of six digit numbers with exactly two $1$'s, two $2$'s, and two $3$'s is 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{0!2!} = \frac{6!}{2!2!2!}$$
The number 
$$\binom{6}{2,2,2} = \frac{6!}{2!2!2!}$$
is a multinomial coefficient.  The general form is 
$$\binom{n}{n_1,n_2,\ldots, n_k} = \frac{n!}{n_1!n_2! \cdots n_k!}$$
where $n = n_1 + n_2 + \cdots + n_k$.  The binomial coefficient that arose in the example corresponds to the special case in which $k = 2$.  
A: There are 6 slots you can put the first $1$, then 5 slots where you can put the second $1$.  But you can fill these slots in either order, so you must divide that number by 2:


*

*Number of ways to put two $1$s in two out of six slots:  ${6 \cdot 5
   \over 2}$.


Once you've done that, there are four places you can put the first $2$ and 3 slots where you can put the second $2$... and again, in either order:


*

*Number of ways to put two $2$s in the four remaining slots:  $4 \cdot
   3 \over 2$.


There is only one way to put the remaining two $3$s in the remaining two slots.
Therefore:  ${6 \cdot 5 \cdot 4 \cdot 3 \over 2 \cdot 2} = 90$.
For the second question, the fact that the seven-digit phone number begins with a $7$ just means that you have six unfilled slots.  The number of ways to put in three $1$s is ${6 \choose 3} = 20$.  The remaining digits are $3$s, so there are no alternatives there.
A: The two 1s can sit in ${6\choose2}$ ways, the two 2's can sit in the remaining 4 places in ${4\choose2}$ and the two 3's can sit in ${2\choose2}$.Multiply you get 90.
Similary, the answer for second part is ${6\choose3}$.${3\choose3}$.  I am hoping that you know how to use ${n\choose r} = \frac{n!}{r!.(n-r)!}$ formula
${6\choose2} = \frac{6!}{2!.4!} = \frac{6.5.4!}{2.4!} = \frac{6.5}{2}$
A: The two questions are the same. 
You need the total number of rearragements of the number 112233 which is nothing but $ \frac{6!}{2!2!2!} = 90$.
When all the $6$ digits are different, then the answer is $6!$. But here $3$ digits occur twice.  So we need to ignore their internal permutations of $11, 22,$ and $33$. Therefore we divide by $2!$ for each digit that occur twice. 
Just so that it is clear, if $1$ occur thrice and $2$ occur twice and $3$ occurs only once, then the answer would be $ \frac{6!}{3!2!}$.
