Question

Let n! = 1 x 2 x3 x... x n for integer n >1. If p = (1
x 1!) + (2 x 2!) + (3 x 3!) + ... + (10 x 10!), then p+2
when divided by 11! leaves a remainder of:
(a) 10
(b) 0
(c) 7
(d) 1​

Answer 1

Answer:

answer is

Step-by-step explanation:

10

is answer I hope that it is useful to you