LET n!=1*2*3*4....*n for integer n>=1.if p=(1*1!)+(2*2!)+(3*3!)+......(10*10!),then p+2 when divided by 11! leaves a remainder as:

1. 10

2. 0

3. 7

4. 1

• ans=1;
  
  (1*1!)+(2*2!)+(3*3!)+......(n*n!)=((n+1)!)-1;
  so in que
  p=(1*1!)+(2*2!)+(3*3!)+......(10*10!)
  so p=11! -1
  p+2=11!+1
  
  so when divided by 11! leaves a remainder as:1;
• 11 years agoHelpfull: Yes(4) No(0)