Total ordered pairs (n,r)

[juantheron]

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

 

[Office_Shredder]

You might want to start by asking if r=2, what can n be?

 

[juantheron]

To Office_Shredder would you like to explain me in Detail

Thanks

 

[Office_Shredder]

Can you find n such that
$${n \choose 2} = 120$$?

It's a simple quadratic equation when you write it out, it shouldn't be too hard to do.

 

[juantheron]

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

 

[Office_Shredder]

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.?

 

[juantheron]

Thanks Office_Shredder I did not understand how can i solve for cubic equation.

 

[Office_Shredder]

For r=3 you want to solve
$${n \choose 3} = 120$$
which becomes
$$\frac{n(n-1)(n-2)}{3!} = 120$$
equivalently,
$$n(n-1)(n-2) = 720$$

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
$${n \choose 4} = 120$$
and
$${n \choose 5} = 120$$
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)