• Support PF! Buy your school textbooks, materials and every day products Here!

Total ordered pairs (n,r)

  • Thread starter juantheron
  • Start date
  • #1
21
1
Calculation of Total no. of positive integer ordered pairs ##(n,r)## in ##\displaystyle \binom{n}{r} = 120##

My Solution:: Clearly ##\displaystyle \binom{n}{r} = 120 \Rightarrow \binom{120}{1} = \binom{120}{119} = 120##

So ##(n,r) = (120,1)\;\;,(120,119)## are positive integer ordered pairs which satisfy the given equation.

Now we will calculate for other positive integer ordered pairs whether it is exists or not.

So ##\displaystyle \binom{n}{r} = \frac{n!}{r! \cdot (n-r)!} = 2^3 \times 3 \times 5\Rightarrow \frac{n!}{r! .\cdot (n-r)! \cdot 5} = 2^3 \cdot 3##

So Largest prime factors of ##120## is ##5##. So ##\displaystyle n\geq 5##

Now for ##r##. Here ## 1 \leq r < 119## and ##r \leq \frac{n}{2}##

So my Question is How can I calculate other positive ordered pairs.

So please help me

Thanks
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
You might want to start by asking if r=2, what can n be?
 
  • #3
21
1
To Office_Shredder would you like to explain me in Detail

Thanks
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
Can you find n such that
[tex] {n \choose 2} = 120[/tex]?

It's a simple quadratic equation when you write it out, it shouldn't be too hard to do.
 
  • #5
21
1
To Office_Shredder

##\displaystyle \binom{n}{2}= 120\Rightarrow \frac{n.(n-1)}{2} = 120\Rightarrow n^2-n-240 = 0##


after solving we get ## n = -15## and ##n = 16##

So we can say ##\displaystyle \binom{n}{r} = 120## is satisfies for ##(16,2)##

Now How can I calculate other ordered pairs

and can we restrict positive integer value of ##n## and ##r##

Thanks
 
  • #6
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
OK, now can you solve it for r=3? It's a little harder (it's a cubic, so you probably want to take out a calculator and try a bunch of values - note that it will be for some value n smaller than 16!). How about r=4, r=5 etc.?
 
  • #7
21
1
Thanks Office_Shredder I did not understand how can i solve for cubic equation.
 
  • #8
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
For r=3 you want to solve
[tex] {n \choose 3} = 120 [/tex]
which becomes
[tex] \frac{n(n-1)(n-2)}{3!} = 120 [/tex]
equivalently,
[tex] n(n-1)(n-2) = 720 [/tex]

Based on the r=2 case any solution of n has to be less than 16 (why?), so just take out a calculator and try n=1, n=2, etc. through n=15 and see if there are any possibilities (or factor 720 and see if you can write it as n(n-1)(n-2) for some n)

Then you can similarly solve
[tex] {n \choose 4} = 120 [/tex]
and
[tex] {n \choose 5} = 120 [/tex]
and at some point your choices for n will be small enough that you will be able to conclude that there are no more solutions to be found (think about why this is, it's similar to the reason why for r=3 I know I can exclude n > 15)
 

Related Threads on Total ordered pairs (n,r)

  • Last Post
Replies
2
Views
801
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
6
Views
766
Replies
4
Views
5K
Replies
3
Views
13K
  • Last Post
Replies
16
Views
803
Replies
15
Views
901
  • Last Post
Replies
5
Views
2K
Replies
5
Views
1K
Top