How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen? How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen? (Suppose $n, k ≥ 2, n ≥ k$).
My answer is $2\binom{n-1}{k} - \binom{n-2}{k}$. First, I counted sets without $1$, then without $2$, then found their sum and, as the formulas are the same, it's just $2$ multiplied by the formula.  Then I subtracted all sets with both $1$ and $2$, as they was already counted. Is it correct?
 A: Your answer is correct. Here are some alternate, equivalent expressions.

You can also find it like this: select $k$ numbers from $n$ numbers and remove the selections that contain both $1$ and $2$. Such selections contain $k-2$ numbers from the remaining $n-2$ numbers. Thus the answer can be written as$$\binom nk-\binom{n-2}{k-2}$$

Alternatively, consider two cases:

*

*$1$ is selected: Then $2$ is not selected. So we have $\binom{n-2}{k-1}$ selections.

*$1$ is not selected: $2$ may or may not be selected. We have $\binom{n-1}k$ selections.

This gives total selections$$\binom{n-2}{k-1}+\binom{n-1}k$$
You can prove that all $3$ expressions are identical using Pascal's identity.

A suggestion for asking future questions to confirm your answers: explain your logic in a few lines so that we know how you obtained your expression. I solved the question in a different manner and obtained a different looking but numerically equal expression. It took me some effort to infer that your and my expressions are identical. Also use the solution-verification tag.
A: Your answer is incorrect.
Method 1:  We count directly.
Exactly one of the numbers from the subset $\{1, 2\}$ is selected:  We select one of the two numbers from the subset $\{1, 2\}$ and $k - 1$ of the numbers from the $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{1}\binom{n - 2}{k - 1}$$
ways.
Neither number from the subset $\{1, 2\}$ is selected:  Then we must select $k$ numbers from the $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{0}\binom{n - 2}{k} = \binom{n - 2}{k}$$
ways.
Total:  Since the two cases are mutually exclusive and exhaustive, the number of admissible selections is
$$\binom{2}{1}\binom{n - 2}{k - 1} + \binom{n - 2}{k}$$
Method 2:  We use complementary counting.
There are
$$\binom{n}{k}$$
ways to select a subset of $k$ elements from the set $\{1, 2, 3, \ldots, n\}$.
From these, we must subtract those selections which include both the elements $1$ and $2$.  If we select both $1$ and $2$, we must select $k - 2$ elements from the subset $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{2}\binom{n - 2}{k - 2} = \binom{n - 2}{k - 2}$$
ways.
Hence, the number of admissible selections is
$$\binom{n}{k} - \binom{n - 2}{k - 2}$$
Method 3:  We correct your explanation of your formula.
You correctly found that the number of selections of $k$ elements from the set $\{1, 2, 3, 4, \ldots, n\}$ which exclude $1$ is
$$\binom{n - 1}{k}$$
and that the number of selections of $k$ elements from the set $\{1, 2, 3, 4, \ldots, n\}$ which exclude $2$ is also
$$\binom{n - 1}{k}$$
Notice that the $k$-element subsets which have been counted twice are those which exclude both $1$ and $2$.  Since we only want to count them once, we must subtract them from the total.  Since the subset $\{3, 4, 5,\ldots, n\}$ contains $n - 2$ elements, there are indeed
$$\binom{n - 2}{k}$$
such subsets.  Thus, by the Inclusion-Exclusion Principle, there are
$$\binom{2}{1}\binom{n - 1}{k} - \binom{n - 2}{k}$$
admissible selections, but your explanation for this formula did not make sense.
