We're adding combinations, where the general formula is:
#C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#
Let's substitute in values:
ⁿ⁺¹c₅+ⁿ⁺²c₅+ⁿ⁺³c₅+ⁿ⁺⁴c₅
#=>((n+1)!)/((5!)((n+1)-5)!)+((n+2)!)/((5!)((n+2)-5)!)+((n+3)!)/((5!)((n+3)-5)!)+((n+4)!)/((5!)((n+4)-5)!)#
#=>((n+1)!)/((5!)(n-4)!)+((n+2)!)/((5!)(n-3)!)+((n+3)!)/((5!)(n-2)!)+((n+4)!)/((5!)(n-1)!)#
I'm going to simplify each fraction first before trying to combine them. Remember that #n! = n xx (n-1) xx (n-2) xx ... xx 1# and so:
#=>((n+1)(n)(n-1)(n-2)(n-3)(n-4)!)/((5!)(n-4)!)+((n+2)(n+1)(n)(n-1)(n-2)(n-3)!)/((5!)(n-3)!)+((n+3)(n+2)(n+1)(n)(n-1)(n-2)!)/((5!)(n-2)!)+((n+4)(n+3)(n+2)(n+1)(n)(n-1)!)/((5!)(n-1)!)#
#=>((n+1)(n)(n-1)(n-2)(n-3))/(5!)+((n+2)(n+1)(n)(n-1)(n-2))/(5!)+((n+3)(n+2)(n+1)(n)(n-1))/(5!)+((n+4)(n+3)(n+2)(n+1)(n))/(5!)#
Know that #5! = 120#, and so:
#=>1/120((n+1)(n)(n-1)(n-2)(n-3)+(n+2)(n+1)(n)(n-1)(n-2)+(n+3)(n+2)(n+1)(n)(n-1)+(n+4)(n+3)(n+2)(n+1)(n))#
And we can simplify a bit more:
#=>(((n+1)(n))/120)((n-1)(n-2)(n-3)+(n+2)(n-1)(n-2)+(n+3)(n+2)(n-1)+(n+4)(n+3)(n+2))#
#=>(((n+1)(n))/120)((n-1)(n-2)(n-3)+(n+2)(n-1)(n-2)+(n+3)(n+2)(n-1)+(n+4)(n+3)(n+2))#
#=>((n^2+n)/120)((n^3-6n^2+11n-6)+(n^3-n^2-4n+4)+(n^3+4n^2+n-6)+(n^3+9n^2+26n+24)#
#=>((n^2+n)/120)(4n^3+6n^2+34n+16)#
#=>(1/60)(n^2+n)(2n^3+3n^2+17n+8)#
#=>(1/60)(2n^5+5n^4+20n^3+25n^2+8n)#
And now to see if I got the math right. Suppose #n=5#. Taking the original expression, we'd calculate:
#C_(6,5)+C_(7,5)+C_(8,5)+C_(9,5)=6+21+56+126=209#
And from the expression we've found:
#(1/60)(2(5)^5+5(5)^4+20(5)^3+25(5)^2+8(5))#
#(1/60)(6250+3125+2500+625+40)#
#(1/60)(12540)=209color(white)(000)color(green)root#
