Prove that #""^nC_r=""^nC_(n-r)# hence solve #""^8C_4#?
1 Answer
Mar 19, 2018
# ""^8C_4 = 70 #
Explanation:
By definition we have:
# ""^nC_r = (n!)/(r!(n-r)!) #
And so
# ""^nC_(n-r) = (n!)/((n-r)!(n-(n-r))!) #
# \ \ \ \ \ \ \ \ \ \ = (n!)/((n-r)!(n-n+r)!) #
# \ \ \ \ \ \ \ \ \ \ = (n!)/((n-r)!r!) #
# \ \ \ \ \ \ \ \ \ \ = (n!)/(r!(n-r)!) #
# \ \ \ \ \ \ \ \ \ \ = ""^nC_r \ \ \ QED #
We can calculate:
# ""^8C_4 = (8!)/(4!(8-4)!) #
# \ \ \ \ \ \ = (8!)/(4!4!) #
# \ \ \ \ \ \ = (1.2.3.4.5.6.7.8)/((1.2.3.4)(1.2.3.4)) #
# \ \ \ \ \ \ = (cancel(1.2.3.4).5.6.7.8)/((1.2.3.4)cancel(1.2.3.4)) #
# \ \ \ \ \ \ = (5.6.7.8)/(1.2.3.4) #
# \ \ \ \ \ \ = (5. cancel(6).7.8)/(cancel(2.3).4) #
# \ \ \ \ \ \ = (5.7.8)/(4) #
# \ \ \ \ \ \ = 2.5.7 #
# \ \ \ \ \ \ = 70 #
