Problem: For integers k and n such that 1⩽k<n, let Ckn​=k!(n−k)!n!​. Then (k+1n−2k−1​)Ckn​ is an integer
Answer Choices:
A. For all k and n
B. For all even values of k and n, but not for all k and n
C. For all odd values of k and n, but not for all k and n
D. If k=1 or n−1, but not for all odd values of k and n
E. If n is divisible by k, but not for all even values of k and n
Solution:
Since Ckn​ is always an integer the quantity
(k+1n−2k−1​)Ckn​​=(k+1n−k​−1)Ckn​=k!(n−k)!(k+1)n!(n−k)​−Ckn​=(k+1)!(n−k−1)!n!​−Ckn​=Ck+1n​−Ckn​​
is always an integer.