Write $1001$ as a binomial coefficient $\binom n k$ with $n ≤ 20$

Question

I have this question involving the binomial theorem and have the answer but trying to understand properly how to get there.

In the solutions the first step in the solution is to write $1001 = 7\cdot 11\cdot 13$ and therefore $n\ge 13$,this is what I don't understand, is it because to get a factor of $11$ and $7$ in the coefficient formula we therefore need $n$ to be larger or equal to $13$?

Answer 1

For a given $n$, the largest binomial coefficient $\binom nr$ occurs at the centre or peak, i.e. $$P_n=\binom n{\lfloor{n/2}\rfloor}$$ Testing different values of $n$ shows that $P_{12}=924$ and $P_{13}=1716$, so we conclude that $n\geq13$.

You can also arrive at the same conclusion by considering that $$\binom nr=\frac {n^{\underline{r}}}{r!}=\frac{n\cdot (n-1)\cdot (n-2)\cdots (n-r+1)}{1\cdot 2\cdot 3\cdots r}$$ Since $1001=7\cdot 11\cdot 13$, we need at $13$ in the numerator, hence $n\geq 13$.

If $n=13$, then

• $r<13$ so as not to cancel the $13$ in the numerator (and this also means $r>0$)*
• $r<11$ so as not to cancel the $11$ in the numerator (and this also means $r>2$)*
• $r<7$ so as not to cancel the $7$ in the numerator (and this also means $r>6$)*...but this impossible.

Hence $n\neq13$.

If $n=14$, then

• $r<13$ so as not to cancel the $13$ in the numerator (and this also means $r>1$)*
• $r<11$ so as not to cancel the $11$ in the numerator (and this also means $r>3$)*
• we do not need $r<7$ as there is a $7$ in $14$.
• hence $3<r<11$ may be possible.
• trying $r=4$ gives the desired answer

Hence the conclusion is $$1001=\binom {14}4\qquad\left[=\binom {14}{10}\right]\quad\blacksquare$$

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*because $\binom nr=\binom n{n-r}$

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Special Note

It is interesting to note that $$\binom {14}4=1001=\binom {14}{10}\\ \binom {14}5=2002=\binom {14}{9}\\ \binom {14}6=3003=\binom{14}8\\$$