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