Show that you can choose a subset of these binomial coefficients whose gcd is $1$ Consider the set
$$A_m=\left\{m+1,\binom{m+2}{2},\binom{m+3}{3},\ldots \right\}.$$
I stuck to show that there exists a finite subset $B$ of $A_m$ in which
$$\gcd_{b\in B}(b)=1. $$
I always ran from number theory and now in my research on hit problem, I faced this question. Any help is appreciated. 
 A: Suppose that $p^\alpha \mid m+1$ but $p^{\alpha + 1} \nmid m+1$ (i.e. suppose $p^{\alpha}$ is the largest power of $p$ dividing $m+1$).

Lemma: $\gcd(p, {m + p^\alpha \choose p^{\alpha}}) = 1$.

We prove this using Lucas's Theorem, which says that
$${a \choose b} \equiv \prod_{i \leq k} {a_i \choose b_i} \mod{p},$$
where 
$$\begin{align}
a &= a_0 + a_1 p + a_2 p^2 + \cdots + a_k p^k \\
b &= b_0 + b_1 p + b_2 p^2 + \cdots + b_k p^k
\end{align}$$
and $0 \leq a_i, b_i < p$. These are the base $p$ expansions of $a$ and $b$.
Let $m_i$ denote the base $p$ coefficients of $m$.
As $p^\alpha \mid m+1$, we see that $m_0 = m_1 = \cdots = m_{\alpha - 1} = p-1$. To see this, call $c = m+1$, and let $c_i$ denote the base $p$ coefficients of $c$. As $c$ is divisible by $p^\alpha$, we must have $c_0 = \cdots = c_{\alpha - 1} = 0$, from which it follows that $m_0 = \cdots = m_{\alpha - 1} = p-1$.
Further, we note that $m_{\alpha} \neq p-1$, since otherwise $m+1$ would be divisible by $p^{\alpha + 1}$, contradicting the fact that $p^\alpha$ is the largest power of $p$ dividing $m+1$.
Adding $p^\alpha$ changes the coefficient of $p^\alpha$ by $1$.
Letting $A_i$ denote the base $p$ coefficients of $A = m + p^\alpha$, we have shown that $A_i = p-1$ for $0 \leq i < \alpha$ and $1 \leq A_\alpha \leq p-1$.
Letting $B_i$ denote the base $p$ coefficients of $B = p^\alpha$, we see of course that $B_i = 0$ unless $i = \alpha$, in which case $B_\alpha = 1$.
We thus have that
$$\begin{align}
{m + p^\alpha \choose p^\alpha} = {A \choose B} &\equiv \prod_{i \leq \alpha - 1} {A_i \choose B_i} \cdot {A_\alpha \choose B_\alpha} \cdot \prod_{i \geq \alpha + 1} {A_i \choose B_i} \\
&\equiv \prod_{i \leq \alpha - 1} {A_i \choose 0} \cdot {A_\alpha \choose 1} \cdot \prod_{i \geq \alpha + 1} {A_i \choose 0} \\
&\equiv 1 \cdot {A_\alpha \choose 1} \cdot 1.
\end{align}$$
And as shown above, $1 \leq A_\alpha \leq p-1$, so that $p \nmid A_\alpha!$. Thus ${A_\alpha \choose 1} \not \equiv 0 \pmod p$.
Thus the product is nonzero mod $p$, or equivalently $p$ does not divide ${m + p^\alpha \choose p^\alpha}$. This proves the lemma. $\diamondsuit$

Theorem: There is a finite subset $B$ of $A_m$ such that $\gcd(\{b \in B\}) = 1$.

Let $m+1 = \prod_{i \in I} p_i^{\alpha_i}$ denote the prime factorization of $m+1$. Then in fact
$$ \gcd\left(m+1, { m + p_i^{\alpha_i} \choose p_i^{\alpha_i} }_{i \in I} \right) = 1,$$
where
$${ m + p_i^{\alpha_i} \choose p_i^{\alpha_i} }_{i \in I}$$
is meant to indicate the set of binomial coefficients from each (maximal) prime power occurring in the prime factorization of $m+1$. This is verified one prime at a time by iteratively applying the lemma. $\spadesuit$
