Question #32a4f

1 Answer
Nov 28, 2017

Answer:

#c(5,2)=50#

Explanation:

Usually, notation #c(n,k)# is used for the absolute value for Stirling numbers of the first kind.
https://en.wikipedia.org/wiki/Stirling_number

[Definition]
Stirling number of the first kind #s(n,k)# is the coefficient of #x^k# in the falling factorial
#x(x-1)(x-2)・・・(x-n+1)#

[Calculate c(5,2)]
Therefore, #s(5,2)# is the coefficient of #x^2# in #x(x-1)(x-2)(x-3)(x-4)# and #c(5,2)=abs(s(5,2))#.

#x(x-1)(x-2)(x-3)(x-4)#
#=x(x-1)(x-4)(x-2)(x-3)#
#=x(x^2-5x+4)(x^2-5x+6)#
#=x{(x^2-5x)^2+10(x^2-5x)+24}#
#=x(x^4-10x^3+25x^2+10x^2-50x+24)#
#=x^5-10x^4+35x^3-50x^2+24x#

Thus, #s(5,2)=-50# and #c(5,2)=50#.