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Ok, so I stumbled upon the question on the title these days, when going over Apostol's Calculus I. Now, because of the placement of the question in the exercises section, I'm convinced that the book wants the reader to use the well known identity $${n \choose l} = {n \choose n-l}$$ So, I thought about the following: if we take $l = k$, because of the above identity and data in the problem, we would get $${n \choose 14 - k} = {n \choose k - 4}$$ Now we could naively take the following step: equating $14 - k = k - 4$ yelds $k = 9$, which is the right answer.

Now, I can convince myself that the equality $14 - k = k - 4$ must hold thinking, for example, in Pascal's triangle, which is symmetrical and has equal entries in a given line precisely for $l$ and $n-l$. However by this point the reader, which is supposed to be meeting the subject for the first time, hasn't been introduced to nothing but the (purely algebraic - no combinatorial arguments allowed!) definition of binomial coefficent and the identity above. So, is there any easy way to see that imposing that equality is valid?

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  • $\begingroup$ I'm not sure which definition of the binomial coefficient has been introduced to the reader, but ${n \choose k} = \dfrac{n!}{k!(n-k)!}$ can be easily symmetric in $k$, in the sense that ${n\choose k} = {n \choose n-k}$ $\endgroup$ – taninamdar Oct 18 '14 at 4:03
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    $\begingroup$ Nothing naive about your step. It IS the right answer... $\endgroup$ – imranfat Oct 18 '14 at 4:03
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$$\binom{14}{k} = \binom{14}{k-4}$$ implies that $$\frac{14!}{k! \, (14-k)!} = \frac{14!}{(k-4)! \, (18-k)!},$$ hence $$\frac{k!}{(k-4)!} = \frac{(18-k)!}{(14-k)!}.$$ Now note $$\frac{k!}{(k-4)!} = k(k-1)(k-2)(k-3)$$ and similarly, $$\frac{(18-k)!}{(14-k)!} = (18-k)(17-k)(16-k)(15-k).$$ By inspection, what integer value of $k$ would make these two products equal?


Alternatively, because of the identity $$\binom{n}{k} = \binom{n}{n-k},$$ it follows that two binomial coefficients are equal if their upper indices are the same and their lower indices add up to the upper: i.e., $$\binom{n}{k} = \binom{n}{m}$$ if $k+m = n$. Thus we simply require $k + (k-4) = 14$.

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    $\begingroup$ For sake of completeness, the other two roots of $k(k-1)(k-2)(k-3) = (18-k)(17-k)(16-k)(15-k)$ are $k = 9 \pm i\sqrt{55}$, which are non-real. $\endgroup$ – JimmyK4542 Oct 18 '14 at 6:19
  • $\begingroup$ At first I thought the approach of opening the algebraic formula would be just a mess, but I forgot that in integers problems it's often easy to see what is the only reasonable solution by inspection, and i guess that's exactly what the author wanted one to notice. Thanks ;) $\endgroup$ – ulilaka Oct 18 '14 at 15:48
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$$ {14 \choose k} = {14 \choose k-4} = {14 \choose 14 - (18-k)} $$ So $k = 18-k$ and you get the answer.

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A symmetrical approach, without using the symmetry identity or factorials: $$\require{cancel}\begin{align} \binom{14}k&=\binom{14}{k-4}\\ \frac{14\cdot 13\cdots (14-(k-1))}{1\cdot 2\cdots k}&=\frac {14\cdot 13\cdots (14-(k-4)+1)}{1\cdot2\cdots (k-4)}\\ \frac{14\cdot 13\cdots (15-k)}{1\cdot 2\cdots k}&=\frac {14\cdot 13\cdots (19-k)}{1\cdot2\cdots (k-4)}\\ \frac{\cancel{14\cdot 13\cdots (19-k)}(18-k)\cdots (15-k)}{\cancel{1\cdot 2\cdots (k-4)}(k-3)\cdots k}&=\frac {\cancel{14\cdot 13\cdots (19-k)}}{\cancel{1\cdot2\cdots (k-4)}}\\ (18-k)(17-k)(16-k)(15-k)&=k(k-1)(k-2)(k-3)\\ 18-k&=k\\ k&=9 \end{align} $$

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