## anonymous 4 years ago 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)

1. campbell_st

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....

2. anonymous

yea but we cant use calculators for the test

3. campbell_st

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

4. anonymous

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

5. campbell_st

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

6. anonymous

ok thank you very much