How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$? The question is:

How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$?

My answer to this question is $63$ but the right answer is $64$. I don't know why it is $64$? I need some assistance.
 A: Judging by the comments, you overlooked the one.
Here is a method that can be generalized:
Each factor of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ has the form $2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}13^{a_6}$, where $a_1, a_2, a_3, a_4, a_5, a_6 \in \{0, 1\}$.  Since there are two possible choices for each of the six exponents, there are $2^6 = 64$ possible factors of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
A: You can obtain the solution as well by omitting the factor 1. One unique whole number is if you multiply all 6 primes. The corresponding sequence is $2,3,5,7,11,13$. So we have one sequence. The corresponding binomial coefficient is $\binom{6}{0}=1$
Then we can remove one of the 6 factors. This can be done with all 6 prime factors. The number of sequences is $\binom{6}{1}=5$.
Next we remove two factors. The resulting sequences are: $ 5,7,11,13\quad 3,7,11,13\quad 3,5,11,13\quad 3,5,7,13\quad 3,5,7,11$
$2,7,11,13\quad 2,5,11,13\quad 2,5,7,13\quad 2,5,7,11\quad 2,3,11,13$
$ 2,3,7,13\quad 2,3,7,11\quad 2,3,5,13\quad 2,3,9,11\quad 2,3,7,9$
Basically we two kinds of elements: The removed elements (x) and the remaining (a) elements. Then the number of such sequences is $\binom{6}{2}=15$.
We can go on like this by removing $3, 4$ and $5$ factors. Then the sum is
$$\sum_{i=1}^6 \binom{6}{i}=2^6-1=63$$
The result is easy to evaluate due the binomial theorem, see  here.
One point of view is that $i=6$ is not possible since we would remove all prime factors. On the other hand they don't talk about prime factors, but factors only. Then the number 1 is valid and we obtain $63+1=64$ different numbers.
