Question

find the value of 5p2-4p0+3p1

Answer 1

We know that npr = n!/(n-r)!

Given 5p2 - 4p0 + 3p1

        5p2 = 5!/(5-2)!

                = 5!/3!

                = 5 * 4 * 3 * 2 * 1/3 * 2 * 1

                = 20.


4p0 = 4!/0!

        = 1.


3p1 = 3!/(3-1)!

      = 3 * 2 * 1/2 * 1

      = 3.


Value of 5p2 - 4p0 + 3p1 = 20 - 1 + 3

                                          = 22.



Hope this helps!     

Answer 2

wkt, nPr = n!/(n-r)!

consider 5p2 = 5!/(5-2)! = 5!/3! = (5×4×3×2×1)/(3×2×1) = 5×4 = 20
4p0 = 1 (since nP0 = 1)
3p1 = 3 (since nP1 = n)

5p2-4p0+3p1 = 20 - 1 + 3 = 22

Hope this helps you :)