C r = (n!)/(r!(n-r)!)
9 C 3 = (9!)/(3!*(9-3)!)
9 C 3 = (9!)/(3!*6!)
9 C 3 = (9*8*7*6!)/(3!*6!)
9 C 3 = (9*8*7)/(3!)
9 C 3 = (9*8*7)/(3*2*1)
9 C 3 = (504)/(6) 9 C 3 = 84
Answer: 23
Step-by-step explanation: You can start by setting this up as an equation like so,
X + 3(X) =92 where x is Thomas’ age and the 3x indicates Hullans age being 3 times greater than Thomas.
Combine the x’s
4(x) = 92
Divide by 4 X = 92/4
92/4 = 23, x = 23
This is a geometric sequence with a common ratio of -1/3 and an initial term of -324. Any geometric sequence can be expressed as:
a(n)=ar^(n-1), in this case a=-324 and r=-1/3 so
a(n)=-324(-1/3)^(n-1) so the 5th term will be
a(5)=-324(-1/3)^4
a(5)=-324/81
a(5)= -4
