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math challenge II-A combinatoratics 7.27

 
 
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math challenge II-A combinatoratics 7.27
by Zeyin Wu - Friday, 1 October 2021, 2:49 AM
 

Why the number of ways to choose a,b,ca′,b′,c′ out of 55 is the same as the number to choose a,b,ca,b,c

 out of  7?

 
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Re: math challenge II-A combinatoratics 7.27
by John Lensmire - Friday, 1 October 2021, 12:11 PM
 

Sometimes it's easiest to understand these types of things by listing out examples and comparing.

Start by listing out all possible triples of $(a,b,c)$ such that $1\leq a < b < c\leq 7$ and each differs by more than $1$ from the others. There are not too many, there should be 10 in total. Try to list them from "smallest" to "largest", so start with each of a, b, and c as small as possible and then make them larger.

Then list out all possible triples of $(a', b', c')$ such that $1\leq a'\leq b' \leq c' \leq 5$. Again, there should be 10 in total. Again, try to list them from "smallest" to "largest".

Put these two lists side by side, do you notice any patterns? How do the first numbers relate? How about the second?

This should hopefully help.

Picture of Zeyin Wu
Re: math challenge II-A combinatoratics 7.27
by Zeyin Wu - Friday, 1 October 2021, 10:40 PM
 

Thanks Mr. Lemsmire