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How can I know which of these numbers are the largest? I have to arrange numbers from the largest one to the smallest.

$$\binom{100}{7} \binom{100}{27} \binom{100}{47} \binom{100}{57} \binom{100}{77} \binom{100}{97}$$

Do I need to count everyone? $$\frac{100!}{7!(100-7)!}= \frac{100!}{7!*93!}$$

Like this one? Or is there another way?

Asia
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    The one whose bottom is closest to 100/2 = 50 should be the largest ;) – Theleb Mar 05 '20 at 20:38
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    Use the fact that $$\binom{n}k=\binom{n}{n-k}$$and then$$\binom{n}k\le\binom{n}{k+1}\qquad 0\le k\le\lfloor n/2\rfloor-1$$ – Peter Foreman Mar 05 '20 at 20:40
  • So this inequality works only for k ≥0 and at the same time k≤ (n/2)-1 right? – Asia Mar 05 '20 at 20:55
  • @Asia, The arrangement is: 47, 57,27,7,77,97 as Peter suggested. – sirous Mar 05 '20 at 21:11
  • The binomial coefficient isn't defined or is defined to be equal to zero if $k$ were less than $0$, so to the first part of your question in your comment, yes. To the second part of your question, yes but when $k$ is greater than $n/2$ then the inequality is reversed. So you have for instance $\binom{5}{0}<\binom{5}{1}<\binom{5}{2}$ while you have $\binom{5}{3}>\binom{5}{4}>\binom{5}{5}$... which you should recognize from pascal's triangle as the values $1,5,10,10,5,1$ – JMoravitz Mar 05 '20 at 21:12
  • @sirous fix your mistake. $7$ is farther away from the center than $77$ is. – JMoravitz Mar 05 '20 at 21:14
  • @sirous No, $\binom{100}{7}$ which is equal to $\binom{100}{93}$ is much smaller than $\binom{100}{43}$ which is equal to $\binom{100}{57}$. – JMoravitz Mar 05 '20 at 21:20
  • @J Moravitz, you are right – sirous Mar 05 '20 at 21:21

1 Answers1

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It is well known that the entries in a row of Pascal's triangle increase as one goes towards the center of the row and then decrease as one goes away from the center and that the rows are symmetric.

In "math-heavy" symbolic notation, that is $$\binom{n}{0}<\binom{n}{1}<\binom{n}{2}<\dots<\binom{n}{\lfloor n/2\rfloor}=\binom{n}{\lceil n/2\rceil} >\dots>\binom{n}{n-2}>\binom{n}{n-1}>\binom{n}{n}$$ and

$$\binom{n}{k}=\binom{n}{n-k}$$

So, just arrange them based on whose bottom value is closest to the center value of $50$. $47$ is closest to $50$, followed by $57$ and so on... giving

$$\binom{100}{47}>\binom{100}{57}>\binom{100}{27}>\binom{100}{77}>\binom{100}{7}>\binom{100}{97}$$

JMoravitz
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