What is the value of r if the permutation is 13Pr=156?

Answer 1

13Pr=!3!/(13-r)!=(13x12x11x...x2x1)/{(13-r)x(12-r)x(11-r)x...x(2)x(1)]

Divide 156 by 13, then by 12 then by 11... until you get 1

you get to 1 in two steps, that means r=2

Check:

13!/(13-2)!=13!/11!=156

Answer 2

156 = 13*12

So, you must have 13P2