5

What is the value of $r$ for which $$\binom{30}{r}\binom{20}{0} + \binom{30}{r-1}\binom{20}{1} + \ldots +\binom{30}{0}\binom{20}{r}$$ is maximum?

This is how I interpreted it: The above expression is equivalent to choosing $r$ objects from $50$ objects. So it’s value given by $\binom{50}{r}$. Now $\binom{50}{r}$ is maximum at $r=25$ So the answer should be $25$. But actually, the correct answer is $ 20$. How is that possible? And what is wrong with my reasoning?

Aditi
  • 1,349
  • 2
    How do you define $\binom{20}{25}$? – ajotatxe Dec 16 '17 at 14:30
  • @ajotatxe But is it necessary that the expansion will contain $\binom{20}{25}$ ? Doesn’t that depend on the value of $r$ ? For example what if $r=10$ ? – Aditi Dec 16 '17 at 14:36
  • @Aditi, See also: https://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-frac-n2-rceil – lab bhattacharjee Dec 16 '17 at 14:50
  • 5
    This idea that the Vandermode identity isn't valid if $r > 20$ is just ridiculous. The binomial coefficient $\binom{20}{25}$ is perfectly well defined: it is the number of ways to choose 25 objects from a set of 20; equivalently, it is the coefficient of $x^{25}$ in $(1 + x)^{20}$. The Vandermode identity says that the coefficient of $x^r$ is the same on both sides of $(1 + x)^{20}(1 + x)^{30} = (1 + x)^{50}$. This works for all $r$ (even $r > 50$). The only thing not defined for $r > m$ is the formula $$ \binom{m}{r} = \frac{m!}{r!(m - r)!}. $$ However the left hand side is still defined. – Trevor Gunn Dec 16 '17 at 15:07
  • 2
    Your teacher or textbook needs to stop trying to trick students with such nonsense and go back to teaching mathematics. Especially when they are wrong in the first place. – Trevor Gunn Dec 16 '17 at 15:08
  • @TrevorGunn I’m sorry but I’m finding it difficult to understand how is it possible to choose $25$ objects from a set of $20$ – Aditi Dec 16 '17 at 16:03
  • 1
    @Aditi That's the point. It's definitely (and definedly) not possible. Therefore, $\binom{20}{25} = 0$ makes perfect sense, and almost everybody, except apparently your book, defines it that way. – JiK Dec 16 '17 at 16:36
  • 1
    @Aditi We all agree that it is not possible to choose a subset of $25$ objects out of a set of $20$. Equivalently, there is no way to do it. Equivalently, there are $0$ ways to do it. Equivalently, $\binom{20}{25}=0$. [Unfortunately, it sounds like your book is instead choosing to declare that $\binom{20}{25}$ is just undefined. Of course everyone is free to choose their own conventions; ideally they would also state their choices explicitly so you're not left guessing on questions like these! Anyway: your book's convention is what it is; it's also the less convenient one, in my experience.] – mathmandan Dec 16 '17 at 16:51
  • @mathmandan Alright , thanks for telling me , I didn't know that $\binom{20}{25}$ could be defined. That’s what we were taught I guess so I couldn’t think that way. Thanks anyway :) – Aditi Dec 16 '17 at 17:20

3 Answers3

3

Consider

$$\binom{30}{r}\binom{20}{0} + \binom{30}{r-1}\binom{20}{1} + \ldots +\binom{30}{0}\binom{20}{r} =\binom{50}{r}$$ like you said.

However, notice that the expression is only defined up to $r=20$. We know that $\binom{50}{r}$ is increasing in $r$ for $r<25$. Hence, the maximum value of the expression is that when $r=20$.

Karn Watcharasupat
  • 4,323
  • Ohhhh I get it now thank you very much ! – Aditi Dec 16 '17 at 14:45
  • As was said in the comments to the question, it is usual to extend the definition of binomial coefficients such that $\binom{20}{21}=\binom{20}{22}=\binom{20}{23}=\dots=0$ and also $\binom{20}{-1}=\binom{20}{-2}=\binom{20}{-3}=\dots=0$. With such a convention, the identity you mention is valid for all $r\in\mathbb{Z}$ (we also consider the empty sum to be zero). – Jeppe Stig Nielsen Dec 16 '17 at 21:41
  • @JeppeStigNielsen Would that not mean that $r=20,21,22,23,24,25$ all give a maximum? – Karn Watcharasupat Dec 17 '17 at 01:59
  • I would say no. For those $r$, the left-hand side still gives a valid way to express $\binom{50}{r}$. So the maximum occurs at $r=25$ only (because $\binom{50}{r}$ has its max at $r=25$). – Jeppe Stig Nielsen Dec 17 '17 at 02:10
1

The formula $$\sum_{j=0}^r\binom aj\binom b{r-j}=\binom{a+b}r$$ is valid only if $r\le\min\{a,b\}$.

ajotatxe
  • 65,084
0

This is a long comment.

Given that binomial coefficients whose lower argument exceeds their higher argument are zero but well-defined, Trevor Gunn has a point. I verified the point numerically; the code is below:

using System.Numerics;
BigInteger Binomial(int n, int k) => k < 0 || k > n ? 0 : k == 0 ? 1 : k * 2 > n ? Binomial(n, n-k) : (Binomial(n, k-1) * n)/k;
BigInteger f(int r)
{
    BigInteger result = 0;
    for(int i = 0; i <= r; ++i) result += Binomial(30, i) * Binomial(20, r-i);
    return result;
}
Console.WriteLine(f(20));// 3,347,717,751,371,268
Console.WriteLine(f(25));// 50,335,833,680,826,500
J.G.
  • 115,835