finding n and k

[homeschool girl]

Let [MATH]n[/MATH] and [MATH]k[/MATH] be positive integers such that [MATH]n < 10^6[/MATH] and [MATH]\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}.[/MATH]Enter an ordered pair [MATH](n,k)[/MATH].


no clue how to do this

 

[Deleted member 4993]

Let [MATH]n[/MATH] and [MATH]k[/MATH] be positive integers such that [MATH]n < 10^6[/MATH] and [MATH]\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}.[/MATH]Enter an ordered pair [MATH](n,k)[/MATH].
no clue how to do this

You know:

\(\displaystyle \binom{54}{13} \ = \ \ \frac{54!}{13! * (54-13)!}\)

continue....

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[homeschool girl]

I don't have any work b/c I don't know where to start

 

[yoscar04]

Subhotosh Khan gave you an example. The general form looks like Try to use it to find each one of the terms of your sum.

 

[pka]

Let [MATH]n[/MATH] and [MATH]k[/MATH] be positive integers such that [MATH]n < 10^6[/MATH] and [MATH]\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}.[/MATH]Enter an ordered pair [MATH](n,k)[/MATH].
no clue how to do this

This answers your query.

 

[Dr.Peterson]

Let [MATH]n[/MATH] and [MATH]k[/MATH] be positive integers such that [MATH]n < 10^6[/MATH] and [MATH]\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}.[/MATH]Enter an ordered pair [MATH](n,k)[/MATH].

no clue how to do this

The important question to me is, how were you expected to figure it out? Not knowing whether it was part of a course you are taking (with what content?) or something else (a contest question?), I have no idea. One way to start, if you know nothing of the identity pka referred to, is to experiment with smaller numbers in a similar pattern, most conveniently while looking at a copy of Pascal's triangle. Then you might at least guess the answer, and then think about how you might prove it (depending on how sure you need to be of your answer, and whether you need to show work).

I will say that, although I didn't have the identity on the tip of my tongue, it was clear to me that the problem must be based on some such identity (one that I assumed I'd seen somewhere), and the experimenting I just suggested is exactly what I would do to determine what it is.

Would you like to tell us what the context was, so we can offer better ideas?

 

[Steven G]

Subhotosh Khan gave you an example. The general form looks like View attachment 21206 Try to use it to find each one of the terms of your sum.

All 42 terms? There has to be an easier way!

 

[Deleted member 4993]

All 42 terms? There has to be an easier way!

Since the OP had "no clue" and did not show any work, we do not even know whether the she knew the "meaning" of the equation. I wanted her to experiment with that definition and realize the shorter way (which pka had guided her to go).