Question

Does anyone know how to compute these following combinations?
1. C(6,4)
2. C(7,5)
3. C(9,6)
4. C(8,2) x C(8,6)

Answer 1

the formula is \[nCr = (n!)/((r!)\times(n-r)!)\]
\[6C4 = 6!/((4!)\times(6-4)!)\]

You should have a calculator function that does it for you....

Answer 2

yea but we cant use calculators for the test

Answer 3

its factorial notation (6x5x4x3x2x1)/((4x3x2x1)x(2x1)) cancel common factors and you'll be left with 3 x 5 = 15

Answer 4

why the 2x1 at the end of the equation come from

Answer 5

because the formula says (n-r)! this is 6 - 4)! or (2)!

Answer 6

ok thank you very much