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.