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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 $13$?

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    $\begingroup$ To get $13$ as a factor of the numerator of $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, you need $n\geq 13$. Because $13$ is... $\endgroup$ – Thomas Andrews Oct 18 '15 at 14:47
  • $\begingroup$ @ThomasAndrews Because its prime? Why is the first logical step in this problem writing down 1001 in prime factors? $\endgroup$ – rb20 Oct 18 '15 at 14:56
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    $\begingroup$ Because binomial coefficients are essentially just a big product, and it's natural to ask "What sort of things need to show up in my product?" Prime factors are just one of those things we can quickly think about. $\endgroup$ – pjs36 Oct 18 '15 at 15:02
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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 $$1001=\binom {14}4\qquad\left[=\binom {14}{10}\right]\quad\blacksquare$$


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


Special Note

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

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  • $\begingroup$ Absolutely wonderful thank you very much $\endgroup$ – rb20 Oct 19 '15 at 11:11
  • $\begingroup$ You're most welcome. $\endgroup$ – hypergeometric Oct 19 '15 at 14:02

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