For integers $n,r$, let $\binom nr = \begin {cases} \binom nr & n\ge r\ge 0 \\ 0 & \text{otherwise} \end {cases}$

Question

For integers $n,r$, let $\binom nr = \begin {cases} \binom nr & n\ge r\ge 0 \\ 0 & \text{otherwise} \end {cases}$. Find the maximum value of $k$ for which the sum $\sum_{i=0}^k \binom {10}{i} \binom{15}{k-i} +\sum_{i=0}^{k+1} \binom {12}{i} \binom {13}{k+1-i}$ exists

I haven’t really understood the question and I have no idea how to begin. Can I get an explanation on what this question actually wants?

The binomial coefficients are always defined thanks to the first part of the question, there is no division by 0 or anything like that, and the indices of summation are finite. Shouldn't it always exist? — Joshua Wang, Mar 02 '21 at 16:27
@JoshuaWang I have absolutely no idea, which is why I asked it on MSE — Aditya, Mar 02 '21 at 16:46
@Aditya: Did you appear in JEE MAINS in which this question was asked, and attempted it? If you did, I would request you to challenge them as their statement is ambiguous and ask for bonus of this question by today (as I think today is the last day of challenging). I have already challenged it. — V.G, Mar 03 '21 at 01:16
@LightYagami they probably won’t give marks because it was an optional question — Aditya, Mar 03 '21 at 02:17
@Aditya: I know that it was optional, but if you appeared in that paper, I sincerely request you to do it because it doesn't matter if it was optional, bonus marks would be given to all, one may also have left it thinking that it is ambiguous...if more number of people challenge them about this, it will definitely have an effect. — V.G, Mar 03 '21 at 02:36
@LightYagami I will challenge it, but they never give marks. Just saying — Aditya, Mar 03 '21 at 03:13
@Aditya: I think they will, because there are a number of mistakes committed by them in other papers as well, so many students are challenging. I hope it favours us, and thank you for accepting my request. — V.G, Mar 03 '21 at 03:21
@LightYagami even in questions that aren’t optional, they don’t give marks. Only for those who attempt it — Aditya, Mar 03 '21 at 03:28
@Aditya: How do you know? If a question gets bonused, IMO marks are given to everyone because one may give the reason that he didn't attempt it because he was finding the statement to be confusing and so to avoid negative marking, he left it. One may also say that, I wasted so much time on that question and ultimately left and later find that question itself was wrong. So, his point is clearly valid, he wasted his precious time on that question, even though he didn't attempt it, so he deserves marks for it. — V.G, Mar 03 '21 at 03:31

score 1 · Accepted Answer · answered Mar 02 '21 at 20:35

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There is no maximum: that sum exists for all integers $k$, if one follows normal practice and takes $\sum_{i=0}^\ell f(i)$ to be $0$ when $\ell<0$, and it exists for all $k\ge 0$ if one takes such sums to be undefined. In fact

$$\begin{align*} &\sum_{i=0}^k\binom{10}i\binom{15}{k-i}+\sum_{i=0}^{k+1}\binom{12}i\binom{13}{k+1-i}\\ &\qquad=\binom{25}k+\binom{25}{k+1}\\ &\qquad=\binom{26}{k+1} \end{align*}$$

for all integers $k$, where I used the Vandermonde identity for the first step.

answered Mar 02 '21 at 20:35

Brian M. Scott

631,399

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I agree. Actually this question was asked in a national level examination held in India (JEE MAINS), and I really despise the organizers for their carelessness of such questions. I myself appeared in the exam, their intention was, find $k$ such that the sum gets maximum, which occurs when $k=12$. In the exam, I somehow had the feeling that they are talking to maximize the sum, but instead went for their statement, and marked $k=25$ thinking they are not considering $\binom{n}r$ to be defined if $r>n$, but later realized that if that was the case, $k$ must not go beyond $10$!!!. – V.G Mar 03 '21 at 01:00
@LightYagami can I know how you found the max value of the sum ie. $k=12$? – Aditya Mar 03 '21 at 02:23
@Aditya: Maximum of binomial coefficient $\binom{n}{r}$ occurs when $r$ is in the middle, that is, if $n$ is even, it occurs at $r=n/2$, but if $n$ is odd, we have two values which are maximum, that is, $\binom{n}{\frac{n+1}{2}}$ and $\binom{n}{\frac{n-1}{2}}$ because they are the same values because of the fact that $\binom{n}{r}=\binom{n}{n-r}$. You can see this by writing the binomial expansion, coefficients are symmetric at the ends, so it rises till the middle and again starts falling... – V.G Mar 03 '21 at 02:32
@Aditya: You may see this. – V.G Mar 03 '21 at 02:34
@LightYagami I know about that, but i fail to see its relevance here. – Aditya Mar 03 '21 at 03:15
@Aditya: Relevance is, their intention was, for what value of $k$ would the given sum be maximized, so the sum evaluates to $\binom{26}{k+1}$ as shown by Brian, so that implies for it to be maximum, $k+1=13$, so $k=12$, that is the current answer that is given by them in the answer key. – V.G Mar 03 '21 at 03:19
@LightYagami I didn’t understand how Brian got $\binom {26}{k+1}$. Is that identity a part of our syllabus? – Aditya Mar 03 '21 at 03:27
@Aditya: The first step is the Vandermonde identity, and the second is just Pascal’s identity. – Brian M. Scott Mar 03 '21 at 03:30
@Aditya: Yes, it is in taught in the syllabus. Read vandemonde's identity, it is not so difficult to understand. – V.G Mar 03 '21 at 03:32
@LightYagami literally never heard of it before – Aditya Mar 03 '21 at 03:33
@Aditya: It’s a very useful one. The link has several proofs; the combinatorial one is especially simple. – Brian M. Scott Mar 03 '21 at 03:34
@BrianM.Scott Yes I understood the combinatorial proof. Thanks for the help – Aditya Mar 03 '21 at 03:39
@Aditya: You’re welcome. – Brian M. Scott Mar 03 '21 at 03:42

For integers $n,r$, let $\binom nr = \begin {cases} \binom nr & n\ge r\ge 0 \\ 0 & \text{otherwise} \end {cases}$

1 Answers1