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