MHB Total no. of positive integer ordered pairs (n,r)

[juantheron]

Calculation of Total no. of positive integer ordered pairs $(n,r)$ in $\displaystyle \binom{n}{r} = 120$

My Try:: 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

 

[I like Serena]

Calculation of Total no. of positive integer ordered pairs $(n,r)$ in $\displaystyle \binom{n}{r} = 120$

My Try:: 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

Hi jacks! :)

Have you tried to simply enumerate them?
That is, build up Pascal's triangle until you can tell there won't be any matches any more?
You may find there are fewer matches than you might think...

\begin{matrix}
&&&1\\
&&1&&1\\
&1&&2&&1\\
1&&3&&3&&1 \\
&&&\dots
\end{matrix}

 

[juantheron]

Thanks Moderator for Help.

But I did not understand it.

Using Guessing I have found $\displaystyle \binom{5}{1} = \binom{16}{2} = 120$

Would you like to explain me in Detail.

Thanks

 

[kaliprasad]

Thanks Moderator for Help.

But I did not understand it.

Using Guessing I have found $\displaystyle \binom{5}{1} = \binom{16}{2} = 120$

Would you like to explain me in Detail.

Thanks

Hello Jacks,
(nCr) = product of r cosecuive numbers (n-r+1) to n/ r!

r =1 1 number = 120
r =2 , product of 2 consecutive numbers = 240 = 15 * 16 so n = 16 so (16c2)
n =3 product of 3 cosecutive numbers = 120* 6 = 720 ( 8 * 9 * 10 = 720)
so (10c3)
n =4. product of 4 cosecutive = 120 * 24 it can be seen if product esists so on

 

[MarkFL]

kaliprasad has given you a good method. I wanted to present it in a slightly different format. Begin with:

$${n \choose r}=120$$

Using the definition of the binomial coefficient, we may write:

$$\frac{n!}{r!(n-r)!}=120$$

$$\frac{n!}{(n-r)!}=120r!$$

Now keep in mind the identity $${n \choose r}={n \choose n-r}$$, so that when you find an ordered pair solution, this will given you another. Now, let's take them case by case on $r$:

i) $r=1$

$$\frac{n!}{(n-1)!}=120\cdot1!$$

$$\frac{n(n-1)!}{(n-1)!}=120\cdot1$$

$$n=120$$

Hence, we have found: $(n,r)=(120,1),\,(120,119)$

ii) $r=2$

$$\frac{n!}{(n-2)!}=120\cdot2!$$

$$\frac{n(n-1)(n-2)!}{(n-2)!}=120\cdot2$$

$$n(n-1)=240$$

$$(n-16)(n+15)=0$$

Discard the negative root, and we have: $(n,r)=(16,2),\,(16,14)$

Now proceed as long as $r$ is smaller than that which you have previously found. For example, the next $r$ is 3, and since this is smaller than 14, you want to continue.

 

[I like Serena]

Thanks Moderator for Help.

But I did not understand it.

Using Guessing I have found $\displaystyle \binom{5}{1} = \binom{16}{2} = 120$

Would you like to explain me in Detail.

Thanks

Pascal's Triangle gives all binomials.
At the top of the triangle, you have $\binom{0}{0}=1$.
The second row contains $\binom{1}{0}=1$ and $\binom{1}{1}=1$.
Each of the numbers is the sum of the two numbers above it.

Continue the triangle until you have found all occurrences of 120.