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Let n and k be positive integers such that \( \binom{n}{k}:\binom{n}{k + 1} = 4:11.\) Find the smallest possible value of n

 Mar 23, 2020
 #1
avatar+111330 
+1

We can simplify this as follows  :

 

    n!                            n!

_______ *11    =   ______________*   4         

(n - k)! k!               (n - k -1)!(k + 1)!

 

 

    11                       4

________  =      _______________

(n - k)!k!             (n - k - 1)! (k+1)!

 

11 (k + 1)!             4 (n - k)!

________  =       _______

     k!                     (n -k -1)!          

 

11 (k + 1)  =     4 ( n - k)

 

11k + 11  = 4n - 4k

 

15k  + 11  =  4n

 

n =  (15k  +  11)

       __________

                4

n will  be the smallest   when  k   =   3

 

          15(3) + 11            56

n  =    _________ =     _____

                4                     4

 

Then  n =  14

 

Proof

 

C(14,3)          364               4

______  =      ____     =   ____

C(14, 4)         1001             11

 

 

cool cool cool

 Mar 23, 2020
 #2
avatar+24995 
+2

Let \(n\) and \(k\) be positive integers such that  \(\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11\).
Find the smallest possible value of \(n\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dbinom{n}{k}:\dbinom{n}{k + 1} &=& 4:11 \\\\ \dfrac{ \dbinom{n}{k}} { \dbinom{n}{k + 1}} &=& \dfrac{4}{11} \quad | \quad \boxed{ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{ \dbinom{n}{k}} { \dfrac{n-k}{k+1} \dbinom{n}{k} } &=& \dfrac{4}{11} \\\\ \dfrac{ 1} { \dfrac{n-k}{k+1} } &=& \dfrac{4}{11} \\\\ \dfrac{k+1} {n-k} &=& \dfrac{4}{11} \\\\ 11(k+1) &=& 4(n-k) \\ 11k+11 &=& 4n-4k \\ \cdots \\ \mathbf{4n} &=& \mathbf{11+15k} \quad | \quad \text{Diophantine equation} \\ \hline \end{array}\)

 

Euler's method for linear diophantine equations  \(\mathbf{4n =11+15k}\)

\(\begin{array}{|rclrcl|} \hline \mathbf{4n} &=& \mathbf{11+15k} \\ n &=& \dfrac{11+15k}{4} \\ n &=& \dfrac{12-1+16k-k}{4} \\ n &=& \dfrac{12+16k-1-k}{4} \\ n &=& 3+4k-\underbrace{\dfrac{1+k}{4}}_{=a} \\ \mathbf{n} &=& \mathbf{3+4k-a} & a&=& \dfrac{1+k}{4} \\ & & & 4a&=& 1+k \\ & & & \mathbf{k} &=& \mathbf{4a-1} \\ \mathbf{n} &=& \mathbf{3+4(4a-1)-a} \\ n &=& 3+16a-4-a \\ \mathbf{n} &=& \mathbf{-1+15a} \qquad a\in \mathbf{Z}\\ && n_{min},~ \text{if}~ a=1 \\ n_{min} &=& -1+15 \\ \mathbf{n_{min}} &=& \mathbf{14} \\\\ \mathbf{k} &=& \mathbf{4a-1} \quad | \quad a=1 \\ k &=& 4 -1 \\ \mathbf{k} &=& \mathbf{3} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dbinom{14}{3}: \dbinom{14}{4} &=& 4:11 \\ \hline \end{array}\)

 

laugh

 Mar 23, 2020

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