Let \(n\) and \(k\) be positive integers such that\(\binom{n}{k}:\binom{n}{k + 1} = 4:11.\)
Find the smallest possible value of \(n\).
+128408 We can simplify this as
n! ( n - [k + 1])! (k + 1)!
_______ * ________________ rearrange as
(n - k!) k! n!
(n - k - 1) ! (k +1)! k + 1
________ * ______ = _____
(n - k)! k! n - k
And we know that
k+ 1 4
____ = ____ cross-multiply
n - k 11
11(k + 1) = 4 ( n - k)
11k + 11 = 4n - 4k
15 k = 4n - 11
15k + 11
_______ = n
4
n will be smallest when k = 3
So n = 14
Proof
C ( 14, 3) 364 4
________ = ____ = ___
C ( 14, 4) 1001 11
+26367 Let \(n\) and \(k\) be positive integers such that \(\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11\). Find the smallest possible value of \(n\).
See here: https://web2.0calc.com/questions/help_7014#r2
